Free math worksheets

- Recall the continuity property of polynomial functions. - Evaluate the function at the target input. - Compare the limiting value with the actual function value.

1. A polynomial is continuous at every real input, so its limit can be found by direct substitution. 2. \(\lim_{x\to-1}P(x)=(-1)^4-3\cdot(-1)+2=6\). 3. Also \(P(-1)=6\). Since the limit equals the function value, \(P\) is continuous at \(x=-1\).

\(P\) is continuous at \(x=-1\), and both the limit and function value equal \(6\).

- Find the domain of the logarithm before discussing continuity. - Use continuity of a composition on the allowed inputs. - Check whether the boundary input itself belongs to the domain.

1. The logarithm requires \(x-1>0\), so the domain is \((1, \infty)\). 2. The function \(x-1\) is continuous, and \(\ln u\) is continuous for \(u>0\), so their composition is continuous throughout that domain. 3. The input \(x=1\) is excluded because \(\ln(0)\) is undefined.

\(L\) is continuous on \((1, \infty)\).

- Evaluate the two formulas through one-sided limits at the joining point. - Identify which branch supplies the actual value at \(0\). - Continuity requires all three quantities to agree.

1. The left-hand limit is \(e^0=1\). 2. The right-hand limit is \(0+1=1\), so the two-sided limit is \(1\). 3. Because the left branch includes \(0\), \(E(0)=e^0=1\). The limit equals the function value, so \(E\) is continuous at \(0\).

\(E\) is continuous at \(x=0\); both one-sided limits and \(E(0)\) equal \(1\).

- Analyze the inner function before the logarithm. - Check whether the inner output can ever leave the logarithm's domain. - Use the composition continuity rule once the domain condition is settled.

1. The inner function \(x^2+1\) is a polynomial, so it is continuous for all real \(x\). 2. Also \(x^2+1\ge1>0\) for every real \(x\), so its values always lie in the logarithm's domain. 3. Therefore the composition \(\ln(x^2+1)\) is continuous on all real numbers.

\(J\) is continuous on \(\mathbb{R}\).

- Identify the continuity of the numerator and denominator separately. - Apply the quotient continuity rule with its denominator restriction. - Use the excluded input to split the real line into intervals.

1. The sine function and the linear function \(x+2\) are continuous for all real inputs. 2. Their quotient is continuous wherever the denominator is nonzero. 3. Since \(x+2=0\) only at \(x=-2\), \(B\) is continuous on \((-\infty, -2)\) and \((-2, \infty)\).

\((-\infty, -2)\) and \((-2, \infty)\).

- Evaluate the function’s behavior from each side of the joining point. - Use the correct piece for the actual value at \(x=0\). - Continuity requires both one-sided limits and the function value to agree.

1. From the left, \(\lim_{x\to0^-}g(x)=\lim_{x\to0^-}(x^2+1)=1\). 2. Since the second rule applies at \(x=0\), \(g(0)=2-0=2\). The right-hand limit is also \(2\). 3. The one-sided limits are different, and the left-hand limit does not equal the function value. Therefore, the two-sided limit does not exist and \(g\) is discontinuous at \(0\).

\(g\) is discontinuous at \(x=0\) because \(\lim_{x\to0^-}g(x)=1\), while \(\lim_{x\to0^+}g(x)=g(0)=2\).

- Check each joining point separately. - Compare the left-hand limit, right-hand limit, and function value. - Use the interval conditions to select the correct rule at each endpoint.

1. At \(x=2\), the left-hand limit is \(2^2-1=3\). The middle piece gives the right-hand limit and function value \(f(2)=1.5\cdot 2=3\). Therefore, \(f\) is continuous at \(2\). 2. At \(x=4\), the left-hand limit from the middle piece is \(1.5\cdot 4=6\). The final piece gives the right-hand limit and function value \(f(4)=\sqrt{4}+4=6\). Therefore, \(f\) is continuous at \(4\).

The function is continuous at both joining points. The common value is \(3\) at \(x=2\) and \(6\) at \(x=4\).

- Imagine tracing the graph from left to right without lifting your pencil. - Distinguish a corner from a hole or jump. - At each input, compare what the graph approaches from both sides with the actual function value.

1. At \(x=-1\), the left-hand limit is \(2\), while the right-hand limit is \(0\). Since the one-sided limits are different, \(f\) has a jump discontinuity at \(x=-1\). 2. At \(x=1\), the graph follows the same smooth piece from both sides and passes through \((1, 2)\). Thus \(\lim_{x\to1}f(x)=f(1)=2\), so \(f\) is continuous there. 3. At \(x=2\), both pieces meet at \((2, 1.5)\). Therefore, the left-hand limit, right-hand limit, and function value all equal \(1.5\), so \(f\) is continuous there. The corner affects differentiability, not continuity.

a) Not continuous; there is a jump at \(x=-1\). b) Continuous at \(x=1\). c) Continuous at \(x=2\), even though the graph has a corner there.

- A change from a curve to a line does not necessarily cause a discontinuity. - Look for an isolated filled point that does not lie on the value approached by the graph.

1. At \(x=0\), the curve approaches \(0\) from the left, and the horizontal piece approaches \(0\) from the right. The filled point gives \(p(0)=0\). Therefore, \(\lim_{x\to0}p(x)=p(0)=0\), so \(p\) is continuous at \(x=0\). 2. At \(x=2\), the graph approaches \(0\) from both sides, so \(\lim_{x\to2}p(x)=0\). However, the filled point gives \(p(2)=1\). Since the limit and function value are different, \(p\) has a removable discontinuity at \(x=2\).

At \(x=0\), \(p\) is continuous. At \(x=2\), \(p\) is not continuous; the discontinuity is removable.

- Check whether the graph approaches the same point from both sides. - Look for holes or jumps. - A corner does not by itself make a function discontinuous. - Compare the limiting value with the actual function value.

1. At \(x=0\), the two pieces meet at \((0, 2)\). The one-sided limits and \(h(0)\) are all \(2\), so \(h\) is continuous there. The corner does not affect continuity. 2. At \(x=2\), the left-hand limit and function value are \(1\), while the right-hand limit is \(3\). Since the one-sided limits are different, \(h\) has a jump discontinuity at \(x=2\). 3. At \(x=4\), the graph lies on one unbroken line segment, so the limit equals \(h(4)=4\). Therefore, \(h\) is continuous at \(x=4\).

At \(x=0\), \(h\) is continuous. At \(x=2\), \(h\) is not continuous; it has a jump discontinuity. At \(x=4\), \(h\) is continuous.

- Start with the general continuity rule for rational functions. - Find every input that makes the denominator zero. - Use those excluded inputs to split the real line into continuity intervals.

1. A rational function is continuous wherever its denominator is nonzero. 2. The denominator factors as \(x^2-9=(x-3)(x+3)\), so it is zero at \(x=-3\) and \(x=3\). 3. Therefore \(R\) is continuous on \((-\infty, -3)\), \((-3, 3)\), and \((3, \infty)\).

\((-\infty, -3)\), \((-3, 3)\), and \((3, \infty)\).

- Find the real domain before discussing continuity. - Interior points and endpoints use slightly different limit conditions. - At the left endpoint of a domain interval, only nearby domain inputs to the right are available.

1. The square root requires \(x+2\ge0\), so the real domain is \([-2, \infty)\). 2. The function is continuous at every interior point of this domain. 3. At the endpoint, \(\lim_{x\to-2^+}\sqrt{x+2}=0=S(-2)\), so \(S\) is continuous on \([-2, \infty)\) using endpoint continuity.

\(S\) is continuous on \([-2, \infty)\). At \(x=-2\), continuity is checked with the right-hand limit because the domain has nearby points only to the right.

- Solve the inequality that makes the radicand nonnegative. - Use continuity of a composition on interior points of the domain. - At each endpoint, approach only through points that remain in the domain.

1. The radicand must satisfy \(4-x^2\ge0\), so the domain is \([-2, 2]\). 2. On the interior \((-2, 2)\), the polynomial \(4-x^2\) is continuous and positive, and square root is continuous on nonnegative inputs, so the composition is continuous. 3. At \(-2\), the right-hand limit equals \(H(-2)=0\); at \(2\), the left-hand limit equals \(H(2)=0\). Thus \(H\) is continuous on \([-2, 2]\).

\(H\) is continuous on \([-2, 2]\), using right-hand continuity at \(-2\) and left-hand continuity at \(2\).

- Translate continuity of \(f\) and \(g\) into limit statements. - Recall the condition needed for the quotient law. - Finish by comparing the quotient limit with the quotient's point value.

1. Continuity gives \(\lim_{x\to a}f(x)=f(a)\) and \(\lim_{x\to a}g(x)=g(a)\). 2. Since \(g(a)\ne0\), the quotient law applies, so \(\lim_{x\to a}Q(x)=\frac{f(a)}{g(a)}\). 3. This equals \(Q(a)\), so the limit equals the function value and \(Q\) is continuous at \(a\).

\(Q\) is continuous at \(x=a\) because the quotient limit exists and equals \(Q(a)=\frac{f(a)}{g(a)}\), with a nonzero denominator.

- Read the y-value approached from the left and from the right. - Compare the common limiting value with the filled point at \(x=3\). - Continuity does not require the graph to be smooth.

1. From both sides of \(x=3\), the graph approaches the y-value \(0\). 2. The filled point is \((3,0)\), so \(A(3)=0\). 3. Since \(\lim_{x\to3}A(x)=A(3)=0\), the function is continuous at \(x=3\). The corner affects differentiability, not continuity.

\(A\) is continuous at \(x=3\) because both the limit and the function value equal \(0\).

- Separate the special point \(x=0\) from all other inputs. - Use the quotient continuity rule where the denominator is nonzero. - At the special point, compare the known limit with the assigned value.

1. For every \(x\ne0\), \(S(x)=\frac{\sin x}{x}\) is a quotient of continuous functions with nonzero denominator, so it is continuous. 2. At \(x=0\), the standard limit \(\lim_{x\to0}\frac{\sin x}{x}=1\). 3. Since \(S(0)=1\), the limit equals the function value at the only point needing a separate check. Therefore \(S\) is continuous on \(\mathbb{R}\).

\(S\) is continuous for every real \(x\).

- Separate interior continuity from endpoint continuity. - At each endpoint, approach only through inputs that lie inside the interval. - Match each one-sided limit with the function value at that endpoint.

1. Continuity on the interior \((0, 3)\) is already given. 2. At the left endpoint, the right-hand limit equals \(f(0)=2\). 3. At the right endpoint, the left-hand limit equals \(f(3)=-1\). These are the endpoint continuity conditions, so \(f\) is continuous on \([0, 3]\). A left-hand limit at \(0\) and a right-hand limit at \(3\) are not required for continuity on this closed interval.

Yes. \(f\) is continuous on \([0, 3]\) because it is continuous on the interior and satisfies the appropriate one-sided continuity conditions at both endpoints.

- Separate the idea of continuity from the behavior of slope. - Compare the limit at \(0\) with the function value there. - Recall the real domain of the cube-root function.

1. The cube-root function is defined and continuous for every real input. 2. In particular, \(\lim_{x\to0}\sqrt[3]{x}=0=C(0)\). 3. A vertical tangent concerns the derivative, not whether nearby function values approach the function value. Thus \(C\) is continuous on \(\mathbb{R}\).

Yes. \(C\) is continuous at \(0\) and on all real numbers; the vertical tangent does not create a discontinuity.

- Combine the domain condition for a square root with the restriction from division. - Determine whether the boundary input is actually in the function's domain. - Endpoint continuity can only include an endpoint where the function itself is defined.

1. The square root requires \(x-2\ge0\), but the result is also in the denominator, so it must be strictly positive. 2. Therefore \(x>2\), giving domain \((2, \infty)\). 3. On this interval the square-root and reciprocal operations are continuous. At \(x=2\), the denominator is \(0\), so the function is undefined and the endpoint cannot be included.

\(D\) is continuous on \((2, \infty)\).

- Start from what continuity of the inner function tells you about its limit. - Check whether the outer square-root function is continuous at that limiting input. - Compare the composed limit with the composed function value.

1. Continuity of \(f\) gives \(\lim_{x\to a}f(x)=9\). Since this limiting value is positive, \(f(x)\) remains positive for inputs sufficiently close to \(a\). 2. The square-root function is continuous at \(9\), so \(\lim_{x\to a}\sqrt{f(x)}=\sqrt{9}=3\). 3. Also \(G(a)=\sqrt{f(a)}=3\). Therefore the limit equals the function value and \(G\) is continuous at \(a\).

\(G(a)=3\), and \(G\) is continuous at \(x=a\).

- Rewrite tangent in terms of sine and cosine if needed. - Find where the denominator of that quotient is zero in the stated interval. - Use those excluded inputs to divide the interval into maximal continuous pieces.

1. The tangent function is continuous wherever \(\cos x\ne0\). 2. On \([0, 2\pi]\), \(\cos x=0\) at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). 3. Therefore the maximal continuity subintervals are \([0, \frac{\pi}{2})\), \((\frac{\pi}{2}, \frac{3\pi}{2})\), and \((\frac{3\pi}{2}, 2\pi]\).

\([0, \frac{\pi}{2})\), \((\frac{\pi}{2}, \frac{3\pi}{2})\), and \((\frac{3\pi}{2}, 2\pi]\).

- Find the left-hand limit using the first piece. - Evaluate the second piece at the joining point. - Set the two values equal and solve for the parameter.

1. The left-hand limit is \(\lim_{x\to-1^-}\frac{6}{x-2}=\frac{6}{-3}=-2\). 2. The second rule gives the function value and right-hand limit: \(g(-1)=a\cdot 2^2-10=4a-10\). 3. Continuity requires \(4a-10=-2\). 4. Solving gives \(4a=8\), so \(a=2\).

\(a=2\)

- Identify where a rational function can fail to be continuous. - Determine when \(x^2+c\) has no real zeros. - Test a simple positive value of \(c\).

1. A rational function is continuous at every point in its domain, so the only possible discontinuities occur where the denominator is \(0\). 2. The denominator equation is \(x^2+c=0\). If \(c>0\), then \(x^2+c>0\) for every real \(x\), so there are no excluded inputs. 3. For example, \(c=1\) gives \(g_1(x)=\frac{1}{x^2+1}\), which is defined and continuous for every real \(x\). 4. Therefore, the statement is true.

The statement is true. Every \(c>0\) makes \(x^2+c\) nonzero for all real \(x\), so \(g_c\) is continuous on \(\mathbb{R}\).

- Substitute the joining-point value into each neighboring piece. - Notice what happens to \(k(x-5)^2\) at \(x=5\). - Set the two expressions at \(x=8\) equal.

1. At \(x=5\), the first piece gives \(g(5)=k\cdot 0^2+2=2\). Its left-hand limit is also \(2\). 2. The second piece is constantly \(2\), so the right-hand limit at \(5\) is \(2\). Thus, \(g\) is continuous at \(5\) for every \(k\). 3. At \(x=8\), the second piece gives \(g(8)=2\) and the left-hand limit is \(2\). 4. The right-hand limit is \(1.5\cdot 8+d=12+d\). Continuity requires \(12+d=2\), so \(d=-10\).

a) Both one-sided limits and \(g(5)\) equal \(2\) for every \(k\). b) \(d=-10\)

- Select the piece containing each input. - At the threshold, compare the left-hand limit, right-hand limit, and function value. - Interpret continuity as the absence of a sudden jump.

1. The second piece applies at \(115\): \(T(115)=2\cdot(115-95)=40\). 2. The third piece applies at \(125\): \(T(125)=2.20\cdot(125-115)+40=22+40=62\). 3. At \(x=115\), the left-hand limit is \(2\cdot(115-95)=40\). The right-hand limit is \(2.20\cdot(115-115)+40=40\), and \(T(115)=40\). 4. Since all three values agree, \(T\) is continuous at \(115\). A very small change across the threshold does not create a sudden jump in the fee.

a) \(T(115)=\$40\) and \(T(125)=\$62\) b) \(T\) is continuous at \(x=115\), because both one-sided limits and the function value equal \(40\). There is no abrupt fee increase at the threshold.

- Evaluate the neighboring pieces at each joining point. - Use the interval conditions to determine the actual function value. - A jump occurs when the one-sided limits are different.

1. At \(x=-2\), the left-hand limit is \(-2+5=3\). The right-hand limit is \((-2)^2-1=3\), and \(g(-2)=3\). Therefore, \(g\) is continuous at \(-2\). 2. At \(x=3\), the left-hand limit is \(3^2-1=8\). The right-hand limit is \(\frac{18}{3}=6\), while \(g(3)=8\). 3. Since the one-sided limits at \(3\) are different, the two-sided limit does not exist. The function has a jump discontinuity at \(3\).

At \(x=-2\): left-hand limit \(=3\), right-hand limit \(=3\), and \(g(-2)=3\); continuous. At \(x=3\): left-hand limit \(=8\), right-hand limit \(=6\), and \(g(3)=8\); not continuous.

- Write the polynomial in factored form using the given zeros and their multiplicities. - At a continuous joining point, the two pieces must approach the same value. - Determine the value the polynomial must have at \(x=2\). - Substitute the joining point to find the scale factor.

1. A cubic polynomial with a double zero at \(0\) and a simple zero at \(4\) has the form \(p(x)=ax^2(x-4)\). 2. Continuity at \(x=2\) requires \(p(2)=8\). 3. Substitute \(x=2\): \(a\cdot 2^2\cdot(2-4)=8\), so \(-8a=8\). 4. Therefore, \(a=-1\), and \(p(x)=-x^2(x-4)=-x^3+4x^2\).

\(p(x)=-x^2(x-4)\), or equivalently \(p(x)=-x^3+4x^2\)

- Find the value that the left-hand piece approaches at the joining point. - Write a general linear function. - Translate the given zero into an equation. - Use the two conditions to form a system of equations.

1. The left-hand limit at \(x=-2\) is \(0.5\cdot(-2)^2+2\cdot(-2)+1=2-4+1=-1\). 2. Continuity requires \(g(-2)=-1\). 3. Write \(g(x)=mx+b\). The zero at \(x=3\) gives \(3m+b=0\), and continuity gives \(-2m+b=-1\). 4. Subtracting the equations gives \(5m=1\), so \(m=0.2\). Then \(b=-3m=-0.6\). 5. Therefore, \(g(x)=0.2x-0.6\).

\(g(x)=0.2x-0.6\)

- Trace each graph toward \(x=1\) from the left and from the right. - A filled point gives the function value; an open circle is not included. - For continuity, the left-hand limit, right-hand limit, and function value must agree. - Analyze the two panels separately.

1. In panel a), \(\lim_{x\to1^-}f(x)=1\), \(\lim_{x\to1^+}f(x)=-1\), and the filled point shows that \(f(1)=1\). Since the one-sided limits are different, the two-sided limit does not exist. Therefore, \(f\) has a jump discontinuity at \(x=1\). 2. In panel b), the left-hand piece gives \(\lim_{x\to1^-}g(x)=1^2-1=0\), and the right-hand piece gives \(\lim_{x\to1^+}g(x)=-1+1=0\). The pieces meet at \((1, 0)\), so \(g(1)=0\). The two one-sided limits and the function value are equal, so \(g\) is continuous at \(x=1\).

a) \(f\) is not continuous at \(x=1\): \(\lim_{x\to1^-}f(x)=1\), \(\lim_{x\to1^+}f(x)=-1\), and \(f(1)=1\). b) \(g\) is continuous at \(x=1\) because \(\lim_{x\to1^-}g(x)=\lim_{x\to1^+}g(x)=g(1)=0\).

- Trace the graph toward each input from both directions. - Compare the left-hand and right-hand limits. - Then compare the common limit, when it exists, with the function value. - Use the filled point at \(x=1\) to identify \(f(1)\).

1. At \(x=-1\), the left-hand piece approaches \(-1\), and the middle piece also approaches \(-1\). The graph includes the point \((-1, -1)\), so \(f(-1)=-1\). Thus \(\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=f(-1)=-1\), and \(f\) is continuous at \(x=-1\). 2. At \(x=1\), the middle piece approaches \(-1\) from the left, while the right-hand piece approaches \(1\). The filled point gives \(f(1)=1\). Since \(\lim_{x\to1^-}f(x)=-1\ne1=\lim_{x\to1^+}f(x)\), the two-sided limit does not exist. Therefore, \(f\) is not continuous at \(x=1\).

At \(x=-1\): \(\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=f(-1)=-1\), so \(f\) is continuous. At \(x=1\): \(\lim_{x\to1^-}f(x)=-1\), \(\lim_{x\to1^+}f(x)=1\), and \(f(1)=1\), so \(f\) is not continuous.

- Find the function value supplied by the branch that includes \(x=2\). - Determine what the right-hand branch approaches at the joining point. - Continuity requires the two branch values at the join to agree.

1. The first piece gives \(f(2)=2^2-3=1\). 2. The right-hand limit is \(\lim_{x\to2^+}(ax-1)=2a-1\). 3. Continuity at \(x=2\) requires \(2a-1=1\). 4. Solving gives \(a=1\).

\(a=1\)

- Compute the left-hand expression at the boundary using the parameter. - Use the second branch for both the right-hand limit and the function value at \(2\). - Set the two one-sided limits equal and solve for \(a\).

1. The left-hand limit at \(x=2\) is \(2a+1\). 2. The right-hand limit and function value are \(F_a(2)=2^2-a=4-a\). 3. Continuity requires \(2a+1=4-a\), so \(3a=3\) and \(a=1\). Then \(F_a(2)=3\).

\(a=1\), and then \(F_a(2)=3\).

- Start with the nonnegative-radicand condition. - Use the numerator zero and denominator zero as sign-analysis breakpoints. - Apply continuity of a composition only on inputs where the inner function is defined and nonnegative.

1. The radicand must satisfy \(\frac{x-1}{x+2}\ge0\), with \(x\ne-2\). 2. A sign analysis at the critical values \(-2\) and \(1\) gives \(x<-2\) or \(x\ge1\). 3. On this domain the rational inner function is continuous and nonnegative, and square root is continuous on nonnegative inputs. Thus \(F\) is continuous on \((-\infty, -2)\) and \([1, \infty)\).

Domain: \((-\infty, -2)\cup[1, \infty)\). Continuity intervals: \((-\infty, -2)\) and \([1, \infty)\).

- Represent the double zero with a squared factor. - Find the value \(g(-2)\) must have for continuity. - Use a factored cubic so the known zero is built into the expression. - Apply the point condition and the joining-point condition to determine the remaining coefficients.

1. The left-hand piece approaches \(-2+5=3\) at \(x=-2\), so continuity requires \(g(-2)=3\). 2. A cubic with a double zero at \(x=1\) can be written as \(g(x)=(ax+b)(x-1)^2\). 3. Since \(g(0)=2\), \(b\cdot(0-1)^2=2\), so \(b=2\). 4. Use \(g(-2)=3\): \((-2a+2)(-3)^2=3\). Thus, \(9(-2a+2)=3\), so \(-18a=-15\) and \(a=\frac{5}{6}\). 5. Therefore, \(g(x)=\left(\frac{5}{6}x+2\right)(x-1)^2\). 6. Expanding gives \(g(x)=\frac{5}{6}x^3+\frac{1}{3}x^2-\frac{19}{6}x+2\).

\(g(x)=\left(\frac{5}{6}x+2\right)(x-1)^2\), or equivalently \(g(x)=\frac{5}{6}x^3+\frac{1}{3}x^2-\frac{19}{6}x+2\)

- Factor the numerator using the denominator factor \(x-3\). - Continuity requires the point value to equal the two-sided limit. - A vertical asymptote requires unbounded behavior, not merely a mismatched point value.

1. Factor the numerator: \(x^2+(k-3)x-3k=(x-3)(x+k)\). Thus, for \(x\ne3\), \(F(x)=x+k\). 2. Therefore \(\lim_{x\to3}F(x)=3+k\). Setting \(3+k=7\) gives \(k=4\). 3. With \(k=4\), continuity at \(x=3\) requires \(m=\lim_{x\to3}F(x)=7\). 4. If \(m=2\), the finite limit still equals \(7\) but differs from the point value. The discontinuity is removable, and \(x=3\) is not a vertical asymptote.

a) \(k=4\) b) \(m=7\) c) The discontinuity is removable, and \(x=3\) is not a vertical asymptote.

- Keep the value \(f(1)\) separate from the limiting value of \(f(u)\) as \(u\to1\). - Continuity of \(g\) controls how close \(g(x)\) is to \(1\), but ask whether it controls equality with \(1\) at nearby inputs. - Try contrasting an inner function that passes through \(1\) only at the target input with one that stays at \(1\) nearby.

1. The outer function has a removable discontinuity at \(u=1\): nearby values approach \(4\), but the value at \(1\) is \(7\). 2. Choose \(g_1(x)=1+x\). It is continuous at \(0\), equals \(1\) at \(0\), and for every nonzero \(x\), \(g_1(x)\ne1\). Therefore \(f(g_1(x))=4+x^2\) for \(x\ne0\), so its limit is \(4\). 3. Choose \(g_2(x)=1\). It is also continuous at \(0\) and equals \(1\) there, but now \(f(g_2(x))=f(1)=7\) for every \(x\). Its limit is \(7\), not \(4\). 4. Knowing only that \(g(x)\to1\) does not say whether nearby values of \(g(x)\) equal the exceptional input \(1\) repeatedly. Because \(f\) is not continuous at \(1\), that distinction can change the composite limit.

a) No. b) One valid pair is \(g_1(x)=1+x\) and \(g_2(x)=1\). Then \(\lim_{x\to0}f(g_1(x))=4\), while \(\lim_{x\to0}f(g_2(x))=7\). c) The difference is whether nearby outputs of the inner function avoid the exceptional input \(1\) or equal it. Since \(f(1)\ne\lim_{u\to1}f(u)\), that distinction matters for the composite.