- Which factor is the same in the numerator and denominator? - For which inputs is cancellation allowed? - The limit depends on values near the missing input, not at it.

1. For \(x\ne5\), canceling \(x-5\) gives \(f(x)=x+1\). 2. The simplified expression approaches \(5+1=6\), so \(\lim_{x\to5}f(x)=6\).

After cancellation, \(f(x)=x+1\) for \(x\ne5\). The limit is \(6\).

- What happens to a difference when its two terms switch places? - Compare the numerator and denominator without setting \(x=3\). - Does the value at the missing input matter for the limit?

1. Reversing the order of subtraction changes the sign: \(3-x=-(x-3)\). 2. For \(x\ne3\), \(f(x)=-\frac{x-3}{x-3}=-1\). 3. The function approaches \(-1\) as \(x\to3\), even though the original quotient is undefined at \(3\).

\(3-x=-(x-3)\). Thus \(f(x)=-1\) for \(x\ne3\), and the limit is \(-1\).

- What happens when two expressions have the same terms but opposite signs between them? - Simplify the product in the numerator before dividing by \(h\). - A limit depends on nearby allowed inputs, even when the original quotient is undefined at the target.

1. The product in the numerator is \((9+h)-9=h\). 2. For \(h\ne0\), \(F(h)=\frac{h}{h}=1\). 3. Therefore \(\lim_{h\to0}F(h)=1\).

The numerator simplifies to \(h\), so \(F(h)=1\) for \(h\ne0\) and the limit is \(1\).

- Take care with the minus sign before the parentheses in the given fraction. - After combining the numerator, check whether the whole quotient has a factor that can be canceled. - Is a factor of \(x\) left after the cancellation?

1. The combined numerator is \(\frac{-x^2}{1+x^2}\). 2. For \(x\ne0\), dividing by \(x\) gives \(Q(x)=\frac{-x}{1+x^2}\). 3. The remaining numerator approaches \(0\), while the denominator approaches \(1\), so the limit is \(0\).

\(Q(x)=\frac{-x}{1+x^2}\) for \(x\ne0\); \(\lim_{x\to0}Q(x)=0\).

- Divide by the highest power of \(x\). - Compare the leading coefficients.

1. Divide the numerator and denominator by \(x\): \(k(x)=\frac{10+\frac{5}{x}}{2-\frac{4}{x}}\). 2. As \(x\to\infty\) or \(x\to-\infty\), the reciprocal terms approach \(0\). 3. Therefore, \(k(x)\to\frac{10}{2}=5\) in both directions.

\(\lim_{x\to\infty}k(x)=5\) and \(\lim_{x\to-\infty}k(x)=5\)

- Divide by the highest power of \(x\) in the denominator. - Compare the degrees of the numerator and denominator.

1. For part a, divide by \(x^2\): \(f(x)=\frac{6+\frac{4}{x}}{3-\frac{1}{x^2}}\). The reciprocal terms approach \(0\), so the limit is \(\frac{6}{3}=2\) in both directions. 2. For part b, the numerator remains constant while the denominator grows without bound as \(|x|\to\infty\). Therefore, the quotient approaches \(0\) in both directions.

a) \(\lim_{x\to\pm\infty}f(x)=2\) b) \(\lim_{x\to\pm\infty}g(x)=0\)

- Factor each numerator completely. - Cancel only after noting that the original input is excluded. - Substitute into the simplified expression.

1. Factor: \(x^2-16=(x-4)(x+4)\) and \(2x-8=2(x-4)\). For \(x\ne4\), the expression equals \(\frac{x+4}{2}\). Its limit at \(4\) is \(4\). 2. Factor: \(x^2+4x+3=(x+1)(x+3)\). For \(x\ne-1\), the expression equals \(x+3\). Its limit at \(-1\) is \(2\).

1) \(4\) 2) \(2\)

- Divide by or factor out the dominant power of \(x\). - Simplify algebraically before taking the limit whenever possible.

1. Divide by \(x\): \(\frac{7-\frac{2}{x}}{\frac{3}{x}-2}\to\frac{7}{-2}=-\frac{7}{2}\). 2. The denominator grows without bound, so the quotient approaches \(0\). 3. Factor \(x^3\): \(x^2-x^3=x^3\left(\frac{1}{x}-1\right)\). The second factor approaches \(-1\), so the expression approaches \(-\infty\). 4. Simplify: \(\frac{x^2+1}{x}-x=x+\frac{1}{x}-x=\frac{1}{x}\), which approaches \(0\).

1) \(-\frac{7}{2}\) 2) \(0\) 3) \(-\infty\) 4) \(0\)

- What happens when you substitute \(x=3\) directly? - Factor the numerator and denominator. - Is there a common factor that can be canceled? - What does a finite limit imply about the type of discontinuity?

1. Direct substitution gives the indeterminate form \(\frac{0}{0}\). 2. Factor the numerator: \(x^2-9=(x-3)(x+3)\). 3. Factor the denominator: \(2x^2-6x=2x(x-3)\). 4. For \(x\neq 3\), cancel the common factor: \(f(x)=\frac{x+3}{2x}\). 5. Evaluate the limit: \(\lim_{x\to 3}f(x)=\frac{3+3}{2(3)}=1\). 6. Because the limit exists and is finite, \(x=3\) is a removable discontinuity.

\(\lim_{x\to 3}f(x)=1\). The discontinuity at \(x=3\) is removable.

- Look for a way to remove the difference involving the square root. - Use an equivalent expression valid for nearby inputs even if not at the target itself. - After simplifying the indeterminate form, try substitution again.

1. Multiply numerator and denominator by the conjugate \(\sqrt{x}+3\). 2. The numerator becomes \(x-9\), so for \(x\ne9\), the expression simplifies to \(\frac{1}{\sqrt{x}+3}\). 3. Now substitute \(x=9\): \(\frac{1}{3+3}=\frac{1}{6}\).

\(\frac{1}{6}\).

- Recognize the numerator as a standard algebraic factorization. - Simplify the expression for inputs near, but not equal to, the target. - Substitute into the simplified polynomial.

1. Factor \(x^3-8=(x-2)(x^2+2x+4)\). 2. For \(x\ne2\), cancel \(x-2\) to obtain \(x^2+2x+4\). 3. Substituting \(x=2\) gives \(4+4+4=12\).

\(12\).

- The obstacle is the radical difference in the denominator. - Use an equivalent nearby expression that removes the vanishing denominator. - Try direct substitution again after simplification.

1. Multiply numerator and denominator by \(\sqrt{x+1}+1\). 2. The denominator becomes \((x+1)-1=x\), so for \(x\ne0\) the expression simplifies to \(\sqrt{x+1}+1\). 3. Substituting \(x=0\) gives \(1+1=2\).

\(2\).

- Recognize the numerator as a standard cubic factorization. - Remove the factor responsible for the indeterminate form. - Evaluate the remaining polynomial at the target input.

1. Factor \(x^3+1=(x+1)(x^2-x+1)\). 2. For \(x\ne-1\), cancel \(x+1\) to obtain \(x^2-x+1\). 3. Substituting \(x=-1\) gives \(1+1+1=3\).

\(3\).

- Expand the power before trying to cancel anything. - After subtracting the constant term, look for a common factor of \(x\). - Substitute only after simplifying the indeterminate form.

1. Expand \((1+x)^3=1+3x+3x^2+x^3\). 2. Subtracting \(1\) gives \(3x+3x^2+x^3=x(3+3x+x^2)\). 3. For \(x\ne0\), cancel \(x\), then let \(x\to0\) to obtain \(3\).

\(3\).

- Both numerator and denominator have standard factorizations. - Look for the common factor responsible for the indeterminate form. - Evaluate the simplified quotient after canceling that factor.

1. Factor \(x^2-1=(x-1)(x+1)\) and \(x^3-1=(x-1)(x^2+x+1)\). 2. For \(x\ne1\), cancel the common factor \(x-1\). 3. The remaining expression approaches \(\frac{1+1}{1+1+1}=\frac{2}{3}\).

\(\frac{2}{3}\).

- Expand the square completely before simplifying. - Look for a common factor matching the denominator. - Evaluate the simplified expression at the limiting value of \(h\).

1. Expand \((2+h)^2=4+4h+h^2\). 2. Subtracting \(4\) gives \(4h+h^2=h(4+h)\). 3. For \(h\ne0\), cancel \(h\), then let \(h\to0\) to obtain \(4\).

\(4\).

- What does direct substitution tell you about the original quotient? - Can the radical expression be rewritten so the factor causing the zero denominator becomes visible? - Remember that a limit concerns inputs near the target, not just the target itself.

1. Multiply the numerator and denominator by \(\sqrt{2x+7}+3\). 2. The denominator becomes \((2x+7)-9=2(x-1)\), so for \(x\ne1\) the expression simplifies to \(\frac{\sqrt{2x+7}+3}{2}\). 3. As \(x\to1\), the simplified quotient approaches \(\frac{3+3}{2}=3\).

\(3\).

- Simplify the subtraction in the numerator before handling the outer division. - Look for the factor that matches the denominator \(x\). - Evaluate the equivalent nearby expression after cancellation.

1. Rewrite the numerator: \(1-\frac{1}{1+x}=\frac{1+x-1}{1+x}=\frac{x}{1+x}\). 2. For \(x\ne0\), dividing by \(x\) gives \(\frac{1}{1+x}\). 3. Substituting \(x=0\) into the simplified expression gives \(1\).

\(1\).

- Check what direct substitution produces. - Look for a difference of squares in the numerator. - After simplifying for nearby inputs, evaluate the remaining expression.

1. Direct substitution gives the indeterminate form \(\frac{0}{0}\), so simplify the expression. 2. Factor the numerator: \(x^2-9=(x-3)(x+3)\). 3. For \(x\ne3\), cancel the common factor to obtain \(x+3\). 4. Therefore \(\lim_{x\to3}\frac{x^2-9}{x-3}=6\).

\(6\)

- The numerator and denominator both vanish at the target input. - Factor each polynomial completely. - Cancel only the factor that is nonzero for nearby inputs.

1. Direct substitution gives \(\frac{0}{0}\). 2. Factor both polynomials: \(x^2+x-6=(x-2)(x+3)\) and \(x^2-4=(x-2)(x+2)\). 3. For \(x\ne2\), the quotient simplifies to \(\frac{x+3}{x+2}\). 4. Taking the limit gives \(\frac{2+3}{2+2}=\frac{5}{4}\).

\(\frac{5}{4}\)

- Check what happens if you substitute the approach value first. - What algebraic identity might help simplify a difference involving a square root? - A transformed expression can agree with the original at every nearby allowed input, even when it has a value at the target.

1. Direct substitution gives \(\frac{0}{0}\). 2. Multiply numerator and denominator by \(\sqrt{1+4x}+1\). 3. For \(x\ne0\), the quotient simplifies to \(\frac{4x}{x(\sqrt{1+4x}+1)}=\frac{4}{\sqrt{1+4x}+1}\). 4. The limit is \(\frac{4}{1+1}=2\).

\(2\).

- First rewrite the numerator as one fraction. - Compare the factors \(1-x\) and \(x-1\). - Evaluate only after the removable factor has been canceled.

1. Combine the terms in the numerator: \(\frac{1}{x}-1=\frac{1-x}{x}\). 2. Then \(\frac{\frac{1-x}{x}}{x-1}=\frac{1-x}{x(x-1)}\). 3. Since \(1-x=-(x-1)\), the expression simplifies to \(-\frac{1}{x}\) for \(x\ne1\). 4. Therefore the limit is \(-1\).

\(-1\)

- Create a denominator that matches the angle inside the sine function. - Use a new input such as \(u=5x\). - Keep track of the constant factor introduced by the rewrite.

1. Rewrite the expression as \(\frac{5}{3}\cdot\frac{\sin(5x)}{5x}\). 2. As \(x\to0\), the input \(5x\to0\). 3. Apply the special limit \(\lim_{u\to0}\frac{\sin u}{u}=1\). 4. The limit is \(\frac{5}{3}\cdot1=\frac{5}{3}\).

\(\frac{5}{3}\)

- First inspect what direct substitution actually produces in each case. - Do not choose a method merely because you see a radical or a sine function. - If substitution gives an indeterminate form, use the expression's structure to decide the next step.

1. In a), direct substitution gives a defined quotient, so direct substitution is the appropriate first method even though a radical is present. 2. In b), substitution produces \(\frac{0}{0}\), and the numerator can be factored to expose the denominator factor. Use factoring and cancellation. 3. In c), substitution produces \(\frac{0}{0}\) with a difference of square roots. Rationalization is the useful first rewrite. 4. In d), direct substitution gives \(\frac{0}{1}\), so no special trigonometric limit is needed. 5. In e), substitution produces \(\frac{0}{0}\), and rewriting can create the standard \(\frac{\sin u}{u}\) form.

a) direct substitution b) factoring and cancellation c) rationalization d) direct substitution e) standard trigonometric limit

- How can you express a whole number with the same denominator as a nearby fraction? - Be careful with the minus sign in front of the parentheses. - After combining the numerator, look for a factor shared with the denominator of the whole expression.

1. Write \(1=\frac{x+1}{x+1}\), using the same denominator as the first term. 2. Subtract the numerators: \(\frac{1}{x+1}-1=\frac{1-(x+1)}{x+1}=-\frac{x}{x+1}\). 3. For \(x\ne0\), the full quotient equals \(-\frac{1}{x+1}\). This approaches \(-1\) as \(x\to0\).

The numerator is \(-\frac{x}{x+1}\). After canceling \(x\) for \(x\ne0\), the limit is \(-1\).

- Which product involving two square-root expressions would remove the radicals from the numerator? - Remember that a fraction stays unchanged if its numerator and denominator are multiplied by the same nonzero expression. - After multiplying, check whether a factor can be canceled at nearby nonzero inputs.

1. Multiply the numerator and denominator by \(\sqrt{x+16}+4\), which makes a difference-of-squares product in the numerator. 2. The new numerator is \((\sqrt{x+16}-4)(\sqrt{x+16}+4)=(x+16)-16=x\). 3. For \(x\ne0\), the quotient simplifies to \(\frac{1}{\sqrt{x+16}+4}\), whose limit at \(0\) is \(\frac{1}{8}\).

Multiply by \(\frac{\sqrt{x+16}+4}{\sqrt{x+16}+4}\). The numerator becomes \(x\), and the limit is \(\frac{1}{8}\).

- Compare each expression at inputs other than \(3\). - Is there a common factor you can cancel at those inputs? - Must two quotients with the same substitution form have the same limit?

1. For every \(x\ne3\), canceling the common factor gives \(A(x)=1\), so \(\lim_{x\to3}A(x)=1\). 2. For every \(x\ne3\), the second expression simplifies to \(B(x)=x-3\), so \(\lim_{x\to3}B(x)=0\). 3. The same undefined substitution form can lead to different limits. The form \(0/0\) alone does not determine a limit.

\(\lim_{x\to3}A(x)=1\) and \(\lim_{x\to3}B(x)=0\). The form \(0/0\) does not determine a limit.

- Which two square terms are being subtracted in the numerator? - Which factor appears in both the top and bottom? - Keep track of the input where the original fraction is undefined.

1. Factor \(x^2-1=(x-1)(x+1)\). 2. For every \(x\ne1\), the quotient equals \(x+1\). 3. The limit depends on these nearby inputs, so it is \(1+1=2\). Canceling does not define the original quotient at \(1\).

\(2\). Cancellation is valid for \(x\ne1\), which is enough to find the limit.

- Which algebraic identity can remove a square root from a difference? - Check which proposed expression is equal to the fraction at nonzero inputs. - The original expression need not be defined at the target for the limit to exist.

1. Multiply by the conjugate \(\sqrt{x+9}+3\) above and below. 2. The numerator becomes \((x+9)-9=x\), which cancels with the original denominator for \(x\ne0\). 3. Choice B is equivalent near zero. Its limit is \(\frac{1}{3+3}=\frac{1}{6}\).

B. For \(x\ne0\), \(\frac{\sqrt{x+9}-3}{x}=\frac{(x+9)-9}{x(\sqrt{x+9}+3)}=\frac{1}{\sqrt{x+9}+3}\). The limit is \(\frac{1}{6}\).

- Multiplying each smaller fraction to a common denominator keeps the subtraction order. - Compare \(2-x\) with \(x-2\). - Simplify only for allowed inputs near the target.

1. \(\frac{1}{x}-\frac{1}{2}=\frac{2-x}{2x}=-\frac{x-2}{2x}\). 2. For \(x\ne2\), the original quotient equals \(-\frac{1}{2x}\). 3. The limit is \(-\frac{1}{4}\).

The correct numerator is \(\frac{2-x}{2x}\); the limit is \(-\frac{1}{4}\).

- Would looking at the two sine terms separately help? - Compare each sine's angle with the denominator. - Keep track of any factor introduced to match an angle.

1. Split the quotient as \(\frac{\sin(3x)}{x}-\frac{\sin x}{x}\) for \(x\ne0\). 2. Rewrite the first term as \(3\frac{\sin(3x)}{3x}\). Both sine ratios now approach \(1\) as \(x\to0\). 3. The limit is \(3\cdot1-1=2\).

Rewrite as \(3\frac{\sin(3x)}{3x}-\frac{\sin x}{x}\). Both sine ratios approach \(1\), so the limit is \(3-1=2\).

- Check whether the claimed factor multiplies every term in the numerator. - At a nearby input, does Jordan's simplified expression equal the original quotient? - Compare the sign of the remaining denominator on both sides of \(2\).

1. The factor \(x-2\) multiplies only the first term of the numerator, so it cannot cancel across the sum. 2. For \(x\ne2\), the quotient is \(x+\frac{4}{x-2}\). 3. As \(x\to2^-\), the second term approaches \(-\infty\); as \(x\to2^+\), it approaches \(+\infty\). 4. There is no finite two-sided limit.

The cancellation across addition is invalid. The left-hand behavior is \(-\infty\), the right-hand behavior is \(+\infty\), and no finite two-sided limit exists.

- Compare the algebraic structure of the two numerators. - What makes direct substitution fail for each quotient? - Which equivalent forms would be easier to evaluate at the target inputs?

1. For a), factor the quadratic \(x^2+3x-10=(x-2)(x+5)\). The limit is \(2+5=7\). 2. For b), multiply by the conjugate. For \(x\ne0\), this becomes \(\frac{1}{\sqrt{5+x}+\sqrt{5}}\). 3. The second limit is \(\frac{1}{2\sqrt{5}}\). Factoring suits a polynomial difference, while the conjugate suits a radical difference.

a) Factor; limit \(7\). b) Rationalize; limit \(\frac{1}{2\sqrt{5}}\). Factoring removes the common polynomial factor in a); multiplying by the conjugate removes the radical difference in b).

- Simplify the quotient before working with the unknown multiplier. - What is the nearby value approached by the remaining polynomial? - Use the required limit to make an equation for the unknown.

1. Factor \(x^2-9=(x-3)(x+3)\). 2. For \(x\ne3\), \(r(x)=k(x+3)\), so its limit at \(3\) is \(6k\). 3. Solve \(6k=12\) to obtain \(k=2\).

\(k=2\), after canceling the common factor for \(x\ne3\).

- Think about division by a nonzero number as multiplication by its reciprocal. - Look for a factor shared by the numerator and denominator of the whole fraction. - Evaluate only after simplifying at inputs where the quotient is defined.

1. Dividing \(\frac{x}{x+2}\) by nonzero \(x\) gives \(\frac{x}{x(x+2)}\). 2. Canceling \(x\) leaves \(g(x)=\frac{1}{x+2}\) for \(x\ne0\) and \(x\ne-2\). 3. The simplified fraction approaches \(\frac{1}{2}\) as \(x\to0\).

For \(x\ne0\) and \(x\ne-2\), \(g(x)=\frac{1}{x+2}\); the limit is \(\frac{1}{2}\).

- Compare the expressions with the special trigonometric ratio supplied in the question. - Which trigonometric relationships could connect the two area expressions to the given limits? - Think about which region contains which when the polygons fit around the circle.

1. \(A_{i,n}=\pi R^2\frac{\sin(2\pi/n)}{2\pi/n}\). 2. Using \(\tan u=\sin u/\cos u\), \(A_{c,n}=\pi R^2\frac{\sin(\pi/n)}{\pi/n}\frac{1}{\cos(\pi/n)}\). 3. Both angles approach \(0\) as \(n\to\infty\). The stated trigonometric limits give \(\lim_{n\to\infty}A_{i,n}=\lim_{n\to\infty}A_{c,n}=\pi R^2\). 4. Every inscribed polygon fits inside the circle; every circumscribed polygon contains it. Thus \(A_{i,n}<\pi R^2<A_{c,n}\). Both bounds approach \(\pi R^2\), confirming the common limiting area.

a) \(A_{i,n}=\pi R^2\frac{\sin(2\pi/n)}{2\pi/n}\); \(A_{c,n}=\pi R^2\frac{\sin(\pi/n)}{\pi/n}\frac{1}{\cos(\pi/n)}\). b) Both area limits equal \(\pi R^2\). c) The inscribed shape is inside the circle and the circumscribed shape is outside it; both approach the circle's area.

- What must be true of the numerator for the discontinuity to be removable? - Substitute the excluded input into the numerator and determine the required output. - After finding the parameter, factor and simplify before evaluating the limit.

1. For the discontinuity at \(x=4\) to be removable, the numerator must also be zero at \(x=4\). 2. Substitute \(x=4\) into the numerator: \(2(4)^2-4k=0\). 3. Solve: \(32-4k=0\), so \(k=8\). 4. With \(k=8\), \(g(x)=\frac{2x^2-8x}{x-4}=\frac{2x(x-4)}{x-4}\). 5. For \(x\neq 4\), the expression simplifies to \(g(x)=2x\). 6. Therefore, \(\lim_{x\to 4}g(x)=\lim_{x\to 4}2x=8\).

\(k=8\), and \(\lim_{x\to 4}g(x)=8\).

- What condition must the denominator satisfy for \(x=2\) to be excluded from the domain? - What must also be true of the numerator for the discontinuity to be removable? - Use the denominator condition to form an equation for \(a\). - After finding \(a\), factor and simplify before evaluating the limit.

1. For \(x=2\) to be a discontinuity, the denominator must equal zero there: \(2^2+2a-6=0\). 2. Solve \(4+2a-6=0\), which gives \(a=1\). 3. With \(a=1\), factor the denominator: \(x^2+x-6=(x-2)(x+3)\). Factor the numerator: \(x^3-4x=x(x-2)(x+2)\). 4. For \(x\neq 2\), simplify: \(g(x)=\frac{x(x-2)(x+2)}{(x-2)(x+3)}=\frac{x(x+2)}{x+3}\). The factor causing the discontinuity cancels completely, so the discontinuity is removable. 5. Evaluate the limit: \(\lim_{x\to 2}g(x)=\frac{2(2+2)}{2+3}=\frac{8}{5}=1.6\).

\(a=1\), and \(\lim_{x\to 2}g(x)=\frac{8}{5}=1.6\).

- Find the limit in terms of \(a\) using the leading coefficients. - Set that expression equal to the required limit.

1. Divide the numerator and denominator by \(x^2\): \(g(x)=\frac{a+\frac{5}{x}}{3-\frac{2}{x^2}}\). 2. Therefore, \(\lim_{x\to\infty}g(x)=\frac{a}{3}\). 3. Set \(\frac{a}{3}=4\), which gives \(a=12\).

\(a=12\)

- Use the ratio of leading coefficients for part a. - Set the limit equal to the desired asymptote value. - Factor and cancel in part c while retaining the domain restriction.

1. Divide the numerator and denominator of \(f_a\) by \(x\): \(f_a(x)=\frac{a+\frac{6}{x}}{2-\frac{4}{x}}\). Thus, \(L=\frac{a}{2}\). 2. A horizontal asymptote \(y=5\) requires \(\frac{a}{2}=5\), so \(a=10\). 3. For \(x\ne0\), \(g(x)=\frac{x(2x+3)}{x}=2x+3=h(x)\). Therefore, both functions approach \(+\infty\) as \(x\to\infty\), and they agree everywhere that \(g\) is defined.

a) \(L=\frac{a}{2}\) b) \(a=10\) c) Both approach \(+\infty\), and \(g(x)=h(x)\) for \(x\ne0\).

- When can a denominator factor be canceled? - What must be true of the numerator at \(x=5\) for the discontinuity to be removable? - What kind of graph results when the simplified expression is linear?

1. For the discontinuity at \(x=5\) to be removable, the numerator must also equal zero at \(x=5\). 2. Solve \(5^2-5k-10=0\): \(15-5k=0\), so \(k=3\). 3. For \(k=3\), factor the numerator: \(x^2-3x-10=(x-5)(x+2)\). 4. For \(x\neq 5\), simplify: \(f(x)=\frac{(x-5)(x+2)}{x-5}=x+2\). 5. Therefore, \(\lim_{x\to 5}f(x)=5+2=7\). 6. The graph is the line \(y=x+2\) with an open point at \((5, 7)\), because \(x=5\) remains excluded from the original domain.

a) \(k=3\) b) \(\lim_{x\to 5}f(x)=7\) c) The graph is the line \(y=x+2\) with an open point at \((5, 7)\).

- Simplify the complex fraction from the inside out. - Look for a factor that cancels the quantity causing the zero denominator. - Substitute only after the nearby expression has been simplified.

1. Combine the numerator: \(\frac{1}{x+2}-\frac{1}{2}=\frac{2-(x+2)}{2(x+2)}=\frac{-x}{2(x+2)}\). 2. For \(x\ne0\), dividing by \(x\) gives \(-\frac{1}{2(x+2)}\). 3. Substituting \(x=0\) into the simplified expression gives \(-\frac{1}{4}\).

\(-\frac{1}{4}\).

- Start by simplifying only the difference in the numerator. - Look for a factor related to the outer denominator. - Evaluate the simplified nearby expression at the target.

1. Combine the numerator: \(\frac{1}{x}-\frac{1}{3}=\frac{3-x}{3x}=-\frac{x-3}{3x}\). 2. For \(x\ne3\), cancel \(x-3\) to get \(-\frac{1}{3x}\). 3. Substituting \(x=3\) gives \(-\frac{1}{9}\).

\(-\frac{1}{9}\).

- The two radicals appear as a difference, suggesting a conjugate. - Use the difference of squares created by that multiplication. - Cancel only after writing an equivalent expression for nearby nonzero inputs.

1. Multiply by the conjugate \(\sqrt{4+x}+\sqrt{4-x}\) over itself. 2. The numerator becomes \((4+x)-(4-x)=2x\), so for \(x\ne0\) the expression simplifies to \(\frac{2}{\sqrt{4+x}+\sqrt{4-x}}\). 3. Substituting \(x=0\) gives \(\frac{2}{2+2}=\frac{1}{2}\).

\(\frac{1}{2}\).

- Factor each polynomial using a structure suited to its degree. - You only need to expose the common factor causing the \(0/0\) form. - Substitute after canceling that factor.

1. Factor \(x^4-16=(x-2)(x+2)(x^2+4)\) and \(x^3-8=(x-2)(x^2+2x+4)\). 2. For \(x\ne2\), cancel \(x-2\). 3. Substitute \(x=2\): \(\frac{4\cdot 8}{12}=\frac{8}{3}\).

\(\frac{8}{3}\).

- Look for a common binomial after grouping terms in the numerator. - Continue factoring until the denominator appears as a factor. - Evaluate only after the removable factor has been canceled.

1. Group the numerator as \(x^3(x+1)-(x+1)=(x+1)(x^3-1)\). 2. Factor \(x^3-1=(x-1)(x^2+x+1)\), then cancel \(x-1\) for \(x\ne1\). 3. The remaining expression is \((x+1)(x^2+x+1)\), which approaches \(2\cdot 3=6\).

\(6\).

- Write \(|x|\) separately for negative and positive inputs. - Simplify each side before taking its limit. - A two-sided limit requires equal one-sided limits.

1. For \(x<0\), \(|x|=-x\), so \(\frac{|x|+x}{x}=\frac{-x+x}{x}=0\). Thus the left-hand limit is \(0\). 2. For \(x>0\), \(|x|=x\), so \(\frac{|x|+x}{x}=\frac{2x}{x}=2\). Thus the right-hand limit is \(2\). 3. Because the one-sided limits are different, the two-sided limit does not exist.

\(\lim_{x\to0^-}\frac{|x|+x}{x}=0\), \(\lim_{x\to0^+}\frac{|x|+x}{x}=2\), and the two-sided limit does not exist.

- Direct substitution does not settle the quotient. Which trigonometric identities could change its form? - Look for a ratio involving a small angle that has a known limit. - After simplifying, examine each factor separately.

1. Use \(1-\cos x=2\sin^2\left(\frac{x}{2}\right)\). 2. Rewrite the quotient as \(\sin\left(\frac{x}{2}\right)\frac{\sin(x/2)}{x/2}\). 3. As \(x\to0\), the first factor tends to \(0\) and the second tends to \(1\). 4. Therefore the limit is \(0\).

\(0\).

- Replace \(1-\cos x\) with a squared sine expression. - Rewrite the denominator using \(x/2\). - Apply the special sine limit after the algebraic rewrite.

1. Use \(1-\cos x=2\sin^2\left(\frac{x}{2}\right)\). 2. Rewrite the quotient as \(\frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2\). 3. The special trigonometric limit makes the squared factor tend to \(1\). 4. Therefore the limit is \(\frac{1}{2}\).

\(\frac{1}{2}\)

- Which part of the numerator prevents you from simplifying by direct substitution? - Think about algebraic transformations that remove a difference of radicals. - See whether one part of the transformed expression has a familiar trigonometric limit.

1. Multiply numerator and denominator by \(\sqrt{1+\sin x}+1\). 2. This gives \(\frac{\sin x}{x(\sqrt{1+\sin x}+1)}\) for \(x\ne0\). 3. Rewrite it as \(\frac{\sin x}{x}\frac{1}{\sqrt{1+\sin x}+1}\). 4. The factors approach \(1\) and \(\frac{1}{2}\), respectively, so the limit is \(\frac{1}{2}\).

\(\frac{1}{2}\).

- Factor the quadratic denominator before combining the fractions. - Use one common denominator for the two terms. - Look for a factor that cancels after the numerators are combined.

1. Factor \(x^2-1=(x-1)(x+1)\). 2. Use the common denominator \((x-1)(x+1)\): \(\frac{x+1-2}{(x-1)(x+1)}\). 3. The numerator is \(x-1\), so the expression simplifies to \(\frac{1}{x+1}\) for \(x\ne1\). 4. Therefore the limit is \(\frac{1}{2}\).

\(\frac{1}{2}\)

- Each approach direction uses a different part of the definition. - Look for an algebraic change that removes division by a quantity tending to zero in each formula. - Do the two one-sided answers agree?

1. Factoring \(x^2-4=(x-2)(x+2)\) gives the left-hand limit \(4\). 2. Multiplying the right-hand expression by the conjugate gives \(\frac{1}{\sqrt{x+2}+2}\) for \(x\ne2\). 3. The right-hand limit is \(\frac{1}{4}\). 4. The two one-sided limits differ, so the two-sided limit does not exist.

\(\lim_{x\to2^-}f(x)=4\), \(\lim_{x\to2^+}f(x)=\frac{1}{4}\), and \(\lim_{x\to2}f(x)\) does not exist.

- Direct substitution tells you that some rewrite is needed, but not which one. - Inspect the numerator and denominator separately for algebraic structures tied to \(x=3\). - Look for two rewrites that can create the same removable factor.

1. Direct substitution gives \(\frac{0}{0}\), so the original form is indeterminate. 2. Factor the numerator as \((x-3)(x+3)\). 3. Rationalize the denominator by multiplying by \(\frac{\sqrt{x+6}+3}{\sqrt{x+6}+3}\). The denominator product becomes \(x-3\). 4. For \(x\ne3\), cancel \(x-3\), leaving \((x+3)(\sqrt{x+6}+3)\). 5. Letting \(x\to3\) gives \(6(3+3)=36\).

Factor the polynomial numerator and rationalize the radical denominator; the limit is \(36\).

- A single rewrite does not have to finish the entire problem. - Ask what rationalization does to the radical difference. - After simplifying the radical part, look for a quotient related to \(\frac{\sin u}{u}\).

1. Direct substitution gives \(\frac{0}{0}\). 2. Rationalize the numerator: for \(x\ne0\), \(\sqrt{1+x}-1=\frac{x}{\sqrt{1+x}+1}\). 3. The quotient becomes \(\frac{x}{\sin x}\cdot\frac{1}{\sqrt{1+x}+1}\). 4. As \(x\to0\), \(\frac{x}{\sin x}\to1\) by the standard sine limit, while \(\frac{1}{\sqrt{1+x}+1}\to\frac{1}{2}\). 5. Therefore the limit is \(\frac{1}{2}\).

Rationalize the numerator, then use the standard sine limit. The limit is \(\frac{1}{2}\).

- Treat \(\frac{0}{0}\) as a diagnosis of the current form, not a conclusion about the limit. - Check whether the proposed operation acts on the whole problematic structure or only on part of it. - Different algebraic forms call for different rewrites: factoring, a conjugate, or a common denominator. - For the cosine expression, look for an identity created by multiplying by \(1+\cos x\).

1. Nora's conclusion is invalid. The form \(\frac{0}{0}\) signals that a rewrite is needed. Factoring and canceling gives \(L_1=4\). 2. Malik's proposal is suitable. Rationalizing gives \(\frac{1}{\sqrt{1+x}+1}\), so \(L_2=\frac{1}{2}\). 3. Elena's proposal is unsuitable because the entire quotient still has denominator \(x-2\) and direct substitution gives \(\frac{0}{0}\). Combining the reciprocal difference gives \(\frac{2-x}{2x(x-2)}=-\frac{1}{2x}\) for \(x\ne2\), so \(L_3=-\frac{1}{4}\). 4. Theo's proposal is suitable. Multiplying by \(1+\cos x\) gives \(\frac{\sin^2 x}{x^2(1+\cos x)}\). Using the standard sine limit then gives \(L_4=\frac{1}{2}\).

\(L_1=4\): Nora's proposal is unsuitable; factor and cancel. \(L_2=\frac{1}{2}\): Malik's rationalization is suitable. \(L_3=-\frac{1}{4}\): Elena's proposal is unsuitable; combine the reciprocal terms before canceling. \(L_4=\frac{1}{2}\): Theo's conjugate/identity rewrite is suitable.