Free math worksheets

- Test each limit by direct substitution before deciding on a method. - Distinguish an indeterminate quotient from a quotient whose denominator alone approaches zero. - L'Hospital's rule applies directly to the standard indeterminate quotient forms, not to every fraction-shaped limit.

1. a) Direct substitution gives \(\frac{0}{0}\), an indeterminate form, so L'Hospital's rule can be applied immediately. 2. b) Direct substitution gives a nonzero numerator over \(0\), not an indeterminate form, so L'Hospital's rule cannot be applied immediately. 3. c) The numerator and denominator both grow without bound, giving the indeterminate form \(\frac{\infty}{\infty}\), so L'Hospital's rule can be applied immediately. 4. d) Direct substitution gives \(\frac{0}{0}\), so L'Hospital's rule can be applied immediately.

a) \(\frac{0}{0}\); yes b) nonzero over \(0\); no c) \(\frac{\infty}{\infty}\); yes d) \(\frac{0}{0}\); yes

- Verify the form produced by direct substitution before using the rule. - Differentiate the numerator and denominator separately. - Re-evaluate the resulting limit after differentiation.

1. Direct substitution gives \(\frac{0}{0}\), so L'Hospital's rule applies. 2. Differentiate numerator and denominator: \(\lim_{x\to3}\frac{2x}{2x-5}\). 3. Substitute \(x=3\): \(\frac{6}{1}=6\).

\(6\)

- Check the numerator and denominator behavior as \(x\) grows. - Once the indeterminate form is confirmed, differentiate the numerator and denominator separately. - Decide whether the new quotient still depends on \(x\).

1. As \(x\to\infty\), the quotient has the indeterminate form \(\frac{\infty}{\infty}\). 2. Apply L'Hospital's rule: \(\lim_{x\to\infty}\frac{4}{7}=\frac{4}{7}\).

\(\frac{4}{7}\)

- Confirm the indeterminate form before differentiating. - Differentiate the composite sine expression carefully. - Evaluate the new limit at the target input.

1. Direct substitution gives \(\frac{0}{0}\), so L'Hospital's rule applies. 2. Differentiate numerator and denominator: \(\lim_{x\to0}\frac{5\cos(5x)}{3}\). 3. Substitution now gives \(\frac{5}{3}\).

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

- Check the form again after the first application of L'Hospital's rule. - A repeated application is justified only if the resulting quotient is still an eligible indeterminate form. - Stop once direct substitution gives a determinate value.

1. Direct substitution gives \(\frac{0}{0}\), so apply L'Hospital's rule: \(\lim_{x\to0}\frac{e^x-1}{2x}\). 2. The new quotient is still \(\frac{0}{0}\), so apply L'Hospital's rule again: \(\lim_{x\to0}\frac{e^x}{2}\). 3. Substitution gives \(\frac{1}{2}\).

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

- Compare the numerator and denominator behavior before choosing the rule. - Differentiate the logarithm and square-root expressions separately. - Simplify the quotient of derivatives before evaluating its limit.

1. As \(x\to\infty\), both \(\ln x\) and \(\sqrt{x}\) grow without bound, so the quotient has the indeterminate form \(\frac{\infty}{\infty}\). 2. Apply L'Hospital's rule: \(\lim_{x\to\infty}\frac{1/x}{1/(2\sqrt{x})}=\lim_{x\to\infty}\frac{2}{\sqrt{x}}\). 3. The final limit is \(0\).

\(0\)

- L'Hospital's rule does not apply directly to a product. - Rewrite one factor so the expression becomes a quotient with an eligible indeterminate form. - After differentiating, simplify before taking the limit.

1. The product has the indeterminate form \(0\cdot(-\infty)\). Rewrite it as \(\frac{\ln x}{1/x}\). 2. As \(x\to0^+\), this quotient has an infinite-over-infinite indeterminate form, so L'Hospital's rule applies. 3. Differentiate numerator and denominator: \(\lim_{x\to0^+}\frac{1/x}{-1/x^2}=\lim_{x\to0^+}(-x)\). 4. Therefore, the limit is \(0\).

\(0\)

- Always test the original limit by direct substitution before applying L'Hospital's rule. - Ask whether the original quotient has an eligible indeterminate form. - If direct substitution already gives a finite value, use that value instead of differentiating.

1. Direct substitution in the original quotient gives \(\frac{1}{1}\), not an indeterminate form. 2. Therefore, L'Hospital's rule is not applicable to the original limit. Differentiating numerator and denominator changes the problem and does not preserve the limit. 3. Direct substitution gives the correct value \(1\).

The rule is not applicable because the original quotient is not indeterminate. The correct limit is \(1\).

- For a variable base raised to a variable exponent, take a logarithm first. - After taking logarithms, rewrite the exponent-product as a quotient. - Apply L'Hospital's rule only after confirming the transformed quotient is indeterminate. - Exponentiate the limit of the logarithm at the end.

1. Let \(y=(1+2x)^{1/x}\). The expression has the indeterminate form \(1^\infty\). 2. Take natural logarithms: \(\ln y=\frac{\ln(1+2x)}{x}\). 3. As \(x\to0^+\), this quotient has the form \(\frac{0}{0}\). Apply L'Hospital's rule: \(\lim_{x\to0^+}\frac{2/(1+2x)}{1}=2\). 4. Therefore, \(\lim\ln y=2\), so \(\lim y=e^2\).

\(e^2\)