﻿ How to Use L'Hôpital's Rule With Exponent Forms - Visual Explanation with color coded examples - 8 Practice Problems explained step by step with interactive problems, showing all work.

# How to Use L'Hôpital's Rule With Exponent Forms: Practice Problems

Step 1

Evaluate the limit in its current form.

$$\displaystyle\lim_{x\to 0^+} x^x = 0^0$$

This is probably the most basic of all of functions that have an indeterminate exponent form in the limit. A good question to start our practice with.

Step 2

Rewrite the limit using the exponential and natural log functions, and pass the limit into the exponent of $$e$$.

\begin{align*} \lim_{x\to 0^+} \blue{x^x} % & = \lim_{x\to 0^+} e^{% \ln\left(% \blue{x^x} \right) } && \blue u = e^{% \ln\blue u } \6pt] % & = e^{% \lim_{x\to 0^+} \ln\left(% x^x \right)} && \text{Pass the limit into the exponent} \end{align*} Step 3 Focus on just the exponent of e . Simplify using the properties of logarithms and re-evaluate the limit. \begin{align*} \lim_{x\to 0^+} \ln\left(% \red x^{% \blue x } \right) % & = \lim_{x\to 0^+} \blue x\cdot \ln \red x && \text{Property of Logarithms} \\[6pt] % & = \blue 0\cdot \red{% \ln 0 } \\[6pt] % & = \blue 0 \cdot \red{% (-\infty) } \end{align*} Step 4 Rewrite the function so the limit has the \frac \infty \infty form. \begin{align*} \lim_{x\to 0^+} \blue x \cdot \red{% \ln x } % & = \lim_{x\to 0^+} \frac{% \red{% \ln x } } {% \blue{1/x} } \\[6pt] % & = \frac{% \red{% \ln 0 } } {% \blue{1/0} } \\[6pt] % & = \frac{% \red{-\infty} } {% \blue \infty } \end{align*} Step 5 Apply L'Hôpital's rule and re-evaluate the limit. \begin{align*} \lim_{x\to 0^+} \frac{% \ln x } {% 1/x } & = \lim_{x\to 0^+} \frac{% \red{% \frac d {dx} \left(% \ln x \right) } } {% \blue{% \frac d {dx} \left(% x^{-1} \right) } } \\[6pt] % & = \lim_{x\to 0^+} \frac{% \red{1/x} } {% \blue{% -x^{-2} } } \\[6pt] % & = \lim_{x\to 0^+} \frac{% \red{1/x} } {% \blue{% -1/x^2 } } \\[6pt] % & = \lim_{x\to 0^+} \red{% \frac 1 x } \cdot \blue{% \left(% -\frac{x^2} 1 \right) } \\[6pt] % & = \lim_{x\to 0^+} -x \\[6pt] % & = 0 \end{align*} Step 6 Go back to the original limit and evaluate it, using the values we found. \begin{align*} \lim_{x\to 0^+} x^x & = e^{% \blue{% \lim\limits_{x\to 0^+} x\ln x} } && \text{ from Step 3 } \\[6pt] % & = e^{% \blue 0 } && \text{ from Step 5 } \\[6pt] % & = 1 \end{align*} Answer \displaystyle \lim_{x\to 0^+} x^x = 1 For reference, here is the graph of the function with the limit value indicated. Step 1 Evaluate the limit in its current form. \displaystyle\lim_{x\to \infty} \blue x^{% \red{% \arccot x } } % = \blue \infty^{% \red{% \arccot \infty } } % = \blue \infty^{% \red 0 } Step 2 Rewrite the limit using the exponential and natural log functions, and pass the limit into the exponent of e . \begin{align*} \lim_{x\to \infty} \blue{% x^{% \arccot x } } % & = \lim_{x\to \infty} e^{% \ln\left(% \blue{% x^{% \arccot x } } \right) } && \blue u = e^{% \ln\blue u } \\[6pt] % & = e^{% \lim_{x\to \infty} \ln\left(% x^{% \arccot x } \right) } && \text{Passing the limit into the exponent.} \end{align*} Step 3 Focus on just the exponent of e . Simplify using the properties of logarithms and re-evaluate the limit. \begin{align*} \lim_{x\to \infty} \ln\left(% x^{% \arccot x } \right) % & = \lim_{x\to \infty} \blue{% \arccot x } \cdot \red{% \ln x } && \text{Properties of Logarithms} \\[6pt] % & = \blue 0 \cdot \red{% (-\infty) } \end{align*} Step 4 Rewrite the function so the limit has the \frac \infty \infty form. \begin{align*} \lim_{x\to \infty} \blue{% \arccot x } \cdot \red{% \ln x } & = \lim_{x\to \infty} \frac{% \red{% \ln x } } {% \blue{% 1/\arccot x } } \\[6pt] % & = \frac{% \red{% \ln \infty } } {% \blue{% 1/\arccot (\infty) } } \\[6pt] % & = \frac{% \red \infty } {% \blue{1/0} } \\[6pt] % & = \frac{% \red{% -\infty } } {% \blue \infty } \end{align*} Step 5 Use L'Hôpital's rule then re-evaluate the limit. \begin{align*} \lim_{x\to \infty} \frac{% \ln x } {% 1/\arccot x } % & = \lim_{x\to \infty} \frac{% \red{% \frac d {dx} \left(% \ln x \right) } } {% \blue{% \frac d {dx} \left(% \arccot x \right)^{-1} } } \\[6pt] % & = \lim_{x\to \infty} \frac{% \red{1/x} } {% \blue{% -(\arccot x)^{-2} \cdot \frac{-1}{1 + x^2} } } \\[6pt] % & = \lim_{x\to \infty} \red{% \frac 1 x } \cdot\blue{% \frac 1 {% \frac 1 {% (\arccot x)^2 (1 + x^2) } } } && \text{Simplifying} \\[6pt] % & = \lim_{x\to \infty} \frac 1 x \cdot (\arccot x)^2 (1 + x^2) \\[6pt] % & = \lim_{x\to \infty} \frac{1 + x^2} x \cdot (\arccot x)^2 \end{align*} To evaluate the limit efficiently, we're going to first simplify the rational factor and then distribute the inverse cotangent. \begin{align*} \lim_{x\to \infty} \frac{1 + x^2} x \cdot (\arccot x)^2 % & = \lim_{x\to \infty} \left(% \frac 1 x + x \right) \cdot (\arccot x)^2 \\[6pt] % & = \lim_{x\to \infty} \left(% \frac 1 x + x \right) \cdot (\arccot x)^2 \\[6pt] % & = \lim_{x\to \infty} \left(% \frac 1 x \cdot (\arccot x)^2 + x (\arccot x)^2 \right) \\[6pt] % & = \lim_{\blue{x\to \infty}} \left(% \frac 1 {\blue x} \cdot (\arccot \blue x)^2 + \blue x (\arccot \blue x)^2 \right) \\[6pt] % & = \frac 1 {\blue \infty} \cdot (\arccot \blue \infty)^2 + \blue \infty (\arccot \blue \infty)^2 \\[6pt] % & = \blue 0 \cdot \blue 0 + \blue \infty (\blue 0) \\[6pt] % & = \infty \cdot 0 \end{align*} Yup! Another indeterminate. Step 6 We'll need to use L'Hôpital's rule again, but only for the x(\arccot x)^2 part (since the other part went to zero). \begin{align*} \lim_{x\to \infty}x \cdot (\arccot x)^2 % & = \lim_{x\to \infty} \frac{% (\arccot x)^2 } {% 1/x } \\[6pt] % & = \lim_{x\to \infty} \frac{% \blue{% \frac d {dx} \left(% \arccot x \right)^2 } } {% \red{% \frac d {dx} \left(% x^{-1} \right) } } \\[6pt] % & = \lim_{x\to \infty} \frac{% \blue{% 2\arccot x \cdot \frac 1 {1 + x^2} } } {% \red{% -x^{-2} } } \\[6pt] % & = \lim_{x\to \infty} \blue{-2\arccot x\cdot \frac 1 {1+x^2}}\cdot \red{x^2} \\[6pt] % & = \lim_{x\to \infty} -2\arccot x\cdot \frac{x^2}{1+x^2} \\[6pt] % & = -2\arccot \infty \cdot 1 \\[6pt] & = -2(0)\\[6pt] & = 0 \end{align*} Step 7 Evaluate the original limit using the values we've found. \begin{align*} \lim_{x\to \infty} x^{\arccot x} % & = \lim_{x\to \infty} e^{% \blue{% \ln\left(% x^{\arccot x} \right) } } \\[6pt] % & = e^{\blue 0} \\[6pt] % & = 1 \end{align*} Answer \displaystyle \lim_{x\to \infty} x^{\arccot x} = 1 For reference, here is the graph of the function. Step 1 Evaluate the limit in its current form. \displaystyle\lim_{x\to 1^-} x^{% \tan\left(% \frac \pi 2 x \right) } % = 1^{% \tan\left(% \frac \pi 2 \right) } % = 1^\infty Step 2 Rewrite the limit using the exponential and natural log functions, and pass the limit into the exponent of e . \begin{align*} \lim_{x\to 1^-} \blue{% x^{% \tan\left(% \frac \pi 2 x \right) } } % & = \lim_{x\to 1^-} e^{% \ln\left(% \blue{% x^{% \tan\frac \pi 2 x } } \right) } \\[6pt] % & = e^{% \lim_{x\to 1^-} \left[% \ln\left(% \blue{% x^{% \tan\frac \pi 2 x } } \right) \right] } \end{align*} Step 3 Focus on just the exponent of e . Simplify using the properties of logarithms and re-evaluate the limit. \begin{align*} \lim_{x\to 1^-} \ln\left(% \blue{% x^{% \tan \frac \pi 2 x } } \right) % & = \lim_{x\to 1^-} \ln\left(% \blue{% x^{% \tan \frac \pi 2 x } } \right) \\[6pt] % & = \lim_{x\to 1^-} \tan\left(% \frac \pi 2 x \right) \cdot \ln x \\[6pt] % & = \tan\left(% \frac \pi 2 \right) \cdot \ln 1 \\[6pt] % & = \infty \cdot 0 \end{align*} Step 4 Rewrite the function so the limit has the \frac 0 0 form. \begin{align*} \lim_{x\to 1^-} \tan\left(% \frac \pi 2 x \right) \cdot \ln x % & = \lim_{x\to 1^-} \frac{% \ln x } {% 1/ \tan\left(% \frac \pi 2 x \right) } \\[6pt] % & = \lim_{x\to 1^-} \frac{% \ln x } {% \cot\left(% \frac \pi 2 x \right) } \\[6pt] % & = \frac{% \ln 1 } {% \cot\left(% \frac \pi 2 \right) } \\[6pt] % & = \frac 0 0 \end{align*} Step 5 Use L'Hôpital's rule then re-evaluate the limit. \begin{align*} \lim_{x\to 1^-} \frac{\ln x}{\cot\left(\frac \pi 2 x\right)} & = \lim_{x\to 1^-} \frac{\blue{\frac d {dx}\left(\ln x\right)}}{\red{\frac d {dx}\left(\cot\left(\frac \pi 2 x\right)\right)}}\\[6pt] & = \lim_{x\to 1^-} \frac{\blue{1/x}}{\red{-\frac \pi 2\csc^2\left(\frac \pi 2 x\right)}}\\[6pt] & = \lim_{x\to \blue{1^-}} \frac{1/\blue x}{-\frac \pi 2\csc^2\left(\frac \pi 2 \blue x\right)}\\[6pt] & = \frac{1/\blue 1}{-\frac \pi 2\csc^2\left(\frac \pi 2\cdot \blue 1\right)}\\[6pt] & = \frac 1 {-\frac \pi 2\cdot 1}\\[6pt] & = \frac 1 {-\pi/2}\\[6pt] & = -\frac 2 \pi \end{align*} Step 6 Evaluate the original limit using the values we've found. \begin{align*} \lim_{x\to 1^-} x^{\tan\left(\frac \pi 2 x\right)} & = e^{\blue{\lim_{x\to 1^-}\left[\ln\left(x^{\tan\frac \pi 2 x}\right)\right]}}\\[6pt] & = e^{\blue{-\frac 2 \pi}}\\[6pt] & \approx 0.5291 \end{align*} Answer \displaystyle \lim_{x\to 1^-} x^{\tan\left(\frac \pi 2 x\right)} = e^{-2/\pi} For reference, here is the graph of the function. Step 1 Evaluate the limit in its current form. \displaystyle\lim_{x\to \blue{0^+}} \left(1 + \blue x\right)^{1/\blue x} = \left(1 + \blue 0\right)^{1/\blue 0} = 1^\infty Step 2 Rewrite the limit using the exponential and natural log functions, and pass the limit into the exponent of e . \begin{align*} \lim_{x\to 0^+} \blue{\left(1 + x\right)^{1/x}} & = \lim_{x\to 0^+} e^{\ln\blue{\left(1 + x\right)^{1/x}}}\\[6pt] & = e^{\lim_{x\to 0^+} \ln\left(\left(1 + x\right)^{1/x}\right)} \end{align*} Step 3 Focus on just the exponent of e . Simplify using the properties of logarithms and re-evaluate the limit. \begin{align*} \lim_{x\to 0^+} \ln\left(\left(1 + x\right)^{1/x}\right) & = \lim_{x\to 0^+} \frac 1 x\cdot \ln\ \left(1 + x\right)\\[6pt] & = \lim_{x\to \blue{0^+}} \frac{\ln\ \left(1 + \blue x\right)}{\blue x}\\[6pt] & = \frac{\ln\ \left(1 + \blue 0\right)}{\blue 0}\\[6pt] & = \frac 0 0 \end{align*} Step 4 Use L'Hôpital's rule then re-evaluate the limit. \begin{align*} \lim_{x\to 0^+} \frac{\ln\ \left(1 + x\right)} x & = \lim_{x\to 0^+} \frac{\blue{\frac d {dx}\left(\ln (1 + x)\right)}}{\red{\frac d {dx}\left(x\right)}}\\[6pt] & = \lim_{x\to 0^+} \frac{\blue{\frac 1 {1 + x}}}{\red 1}\\[6pt] & = \lim_{x\to \blue{0^+}} \frac 1 {1 + \blue x}\\[6pt] & = \frac 1 {1 + \blue 0}\\[6pt] & = 1 \end{align*} Step 5 Evaluate the original limit using the values we've found. \begin{align*} \lim_{x\to 0^+} \left(1 + x\right)^{1/x} & = e^{\blue{\lim_{x\to 0^+} \ln\left(\left(1 + x\right)^{1/x}\right)}}\\[6pt] & = e^{\blue 1}\\[6pt] & = e \end{align*} Answer \displaystyle \lim_{x\to 0^+} \left(1 + x\right)^{1/x} = e . In fact, this limit is an alternate version of the definition of the number e . For reference, below is the graph of the function and the limit value. Step 1 Evaluate the limit in its current form. \displaystyle\lim_{x\to \blue \infty} \left(1 + \frac 2 {\blue x}\right)^{3\blue x} = \left(1 + \frac 2 {\blue \infty} \right)^{\blue \infty} = \left(1 + \blue 0\right)^{\blue \infty} = 1^\infty Step 2 Rewrite the limit using the exponential and natural log functions, and pass the limit into the exponent of e . \begin{align*} \lim_{x\to \infty} \blue{\left(1 + \frac 2 x\right)^{3x}} & = \lim_{x\to \infty} e^{\ln\blue{\left(1 + \frac 2 x\right)^{3x}}}\\[6pt] & = e^{\lim_{x\to \infty} \ln\left(1 + \frac 2 x\right)^{3x}} \end{align*} Step 3 Focus on just the exponent of e . Simplify using the properties of logarithms and re-evaluate the limit. \begin{align*} \lim_{x\to \infty} \ln\left(% 1 + \frac 2 x \right)^{3x} & = \lim_{% \blue{% x\to \infty } } 3\blue x \cdot \ln\left(% 1 + \frac 2 { \blue x } \right) && \text{Propoerties of Logarithms} \\[6pt] % & = \blue \infty \cdot \ln\left(% 1 + \frac 2 { \blue \infty } \right) \\[6pt] % & = \infty \cdot 0 \end{align*} Step 4 Rewrite the function so the limit has the \frac 0 0 form. \begin{align*} \lim_{x\to \infty} 3\,\blue{x} \cdot \ln\left(% 1 + \frac 2 x \right) % & = 3\lim_{x\to \infty} \blue{x} \cdot \ln\left(% 1 + \frac 2 x \right) \\[6pt] % & = 3\lim_{x\to \infty} \frac 1 {% \blue{% 1/x } } \cdot \ln\left(% 1 + \frac 2 x \right) \\[6pt] % & = 3\lim_{x\to \infty} \frac{% \ln\left(% 1 + \frac 2 x \right) } {% \blue{% 1/x } } \\[6pt] % & = 3\left(\frac{% \ln\left(% 1 + \frac 2 \infty \right) } {% 1/ \infty } \right) \\[6pt] % & = 3\left(\frac{% \ln 1 } { 0 } \right) \\[6pt] % & = \frac 0 0 \end{align*} Step 5 Use L'Hôpital's rule then re-evaluate the limit. \begin{align*} 3\lim_{x\to \infty} \frac{% \ln\left(% 1 + \frac 2 x \right) } {% 1/x } % & = 3\lim_{x\to \infty} \frac{% \blue{% \frac d {dx} \left[% \ln\left(% 1 + 2x^{-1} \right) \right] } } {% \red{% \frac d {dx} \left(% x^{-1} \right) } } \\[6pt] % & = 3\lim_{x\to \infty} \frac{% \blue{% \frac 1 {% 1 + 2x^{-1} } \cdot \left(% -2x^{-2} \right) } } {% \red{% -x^{-2} } } \\[6pt] % & = 3\lim_{x\to \infty} \frac{% \frac 1 {% 1 + 2x^{-1} } \cdot 2 \cdot \cancelred{% x^{-2} } } {% \cancelred{% x^{-2} } } \\[6pt] % & = 3\lim_{x\to \infty} \frac 2 {% 1 + 2x^{-1} } \\[6pt] % & =3\cdot \frac 2 {% 1 + \frac 2 \infty } \\[6pt] % & = \frac 6 {% 1 + 0 } \\[6pt] % & = 6 \end{align*} Step 6 Evaluate the original limit using the values we've found. \begin{align*} \lim_{x\to \infty} \left(% 1 + \frac 2 x \right)^{3x} % & = e^{% \blue{% \lim_{x\to \infty} \ln\left(% 1 + \frac 2 x \right)^{3x} } } && \text{ From Step 2 } \\[6pt] % & = e^{% \blue 6 } && \text{ From Step 5 } \\[6pt] % & \approx 403 \end{align*} Answer \displaystyle \lim_{x\to \infty} \left(% 1 + \frac 2 x \right)^{3x} = e^6 For reference, here is the graph of the function. Step 1 Evaluate the limit in its current form. \displaystyle\lim_{x\to 0^+} \left[\ln\left(\frac 1 {\blue x}\right)\right]^{\red x} = \left[\ln\left(\frac 1 {\blue 0}\right)\right]^{\red 0} = \left[\ln\left(\blue \infty\right)\right]^{\red 0} = \blue{\infty}^{\red 0} Since this is one of the indeterminate exponent forms, we'll need to apply the exponential and natural log functions. Step 2 Rewrite the limit using the exponential and natural log functions, and pass the limit into the exponent of e . Note: Since \frac 1 x = x^{-1} , the function can be written a little more compactly as \left[\ln\left(\frac 1 x\right)\right]^x = \left[\ln\left(x^{-1}\right)\right]^x = \left(-\ln x\right)^x. Using this more compact notation, we apply the exponential and natural log functions as follows. \displaystyle\lim_{x\to 0^+} \blue{\left(-\ln x\right)^x} = \displaystyle\lim_{x\to 0^+} e^{\ln\blue{\left(-\ln x\right)^x}} = e^{\lim_{x\to 0^+} \ln\left(-\ln x\right)^x} Step 3 Focus on just the exponent of e . Simplify using the properties of logarithms and re-evaluate the limit. \begin{align*} \lim_{x\to 0^+} \ln\left(\red{-\ln x}\right)^{\blue x} & = \lim_{x\to 0^+} \blue x\cdot \ln\left(\red{-\ln x}\right) && \text{Properties of Logarithms}\\[6pt] & = \blue 0\cdot \ln\left(\red{-\ln 0}\right)\\[6pt] & = \blue 0\cdot \ln\left(\red{\infty}\right)\\[6pt] & = \blue 0\cdot \red{\infty} \end{align*} Step 4 Rewrite the function so the limit has the \frac \infty \infty form. \begin{align*} \lim_{x\to 0^+} \blue x\cdot \red{\ln\left(-\ln x\right)} & = \lim_{x\to 0^+} \frac{\red{\ln\left(-\ln x\right)}}{\blue{1/x}}\\[6pt] & = \lim_{x\to 0^+} \frac{\red{\ln\left(-\ln 0\right)}}{\blue{1/0}}\\[6pt] & = \lim_{x\to 0^+} \frac{\red \infty}{\blue \infty} \end{align*} Step 5 Use L'Hôpital's rule then re-evaluate the limit. \begin{align*} \lim_{x\to 0^+} \frac{\ln\left(-\ln x\right)}{1/x} & = \lim_{x\to 0^+} \frac{\red{\frac d {dx}\left[\ln\left(-\ln x\right)\right]}}{\blue{\frac d {dx} \left(x^{-1}\right)}}\\[6pt] & = \lim_{x\to 0^+} \frac{\red{\frac 1 {-\ln x}\cdot \left(-\frac 1 x\right)}}{\blue{-x^{-2}}}\\[6pt] & = \lim_{x\to 0^+} \red{\frac 1 {x\ln x}}\cdot \blue{\frac 1 {-1/x^2}} && \text{Simplifying}\\[6pt] & = \lim_{x\to 0^+} \frac x {\ln x}\\[6pt] & = \frac 0 {-\infty} && \text{Re-evaluate the limit}\\[6pt] & = 0 \end{align*} Step 6 Evaluate the original limit using the values we've found. \begin{align*} \lim_{x\to 0^+} \left[\ln\left(\frac 1 x\right)\right]^x & = e^{\blue{\lim_{x\to 0^+} \ln\left(-\ln x\right)^x}} && \text{from Step 2}\\[6pt] & = e^{\blue 0} && \text{from Step 5}\\[6pt] & = 1 \end{align*} Answer \displaystyle \lim = \lim_{x\to 0^+} \left[\ln\left(\frac 1 x\right)\right]^x = 1 For reference, the graph of the function and the limit value are shown below. Step 1 Evaluate the limit in its current form. \displaystyle\lim_{% \blue{% x\to \infty } } \left(% \sech \blue x \right)^{% 1/\ln\blue x } % = \left(% \sech \blue \infty \right)^{% 1/\ln\blue \infty } % = 0^{% 1/\infty } % = 0^0 Step 2 Rewrite the limit using the exponential and natural log functions, and pass the limit into the exponent of e . \begin{align*} \lim_{x\to \infty} \blue{% \left(% \sech x \right)^{% 1/\ln x } } % & = \lim_{x\to \infty} e^{% \blue{% \ln\left(% \sech x \right)^{% 1/\ln x } } } % = e^{% \lim_{x\to \infty} \ln\left(% \sech x \right)^{% 1/\ln x } } \end{align*} Step 3 Focus on just the exponent of e . Simplify using the properties of logarithms and re-evaluate the limit. \begin{align*} \lim_{x\to \infty} \ln\left(% \sech x \right)^{% 1/\ln x } % & = \lim_{x\to \infty} \frac 1 {\ln x} \cdot \ln\left(% \sech x \right) && \text{Properties of Logarithms} \\[6pt] % & = \lim_{x\to \infty} \frac{% \ln\left(% \sech x \right) } {\ln x} \\[6pt] % & = \frac{% \ln\left(% \sech \infty \right) } {\ln \infty} \\[6pt] % & = \frac{% \ln 0 } \infty \\[6pt] % & = \frac \infty \infty \end{align*} Step 4 Use L'Hôpital's rule then re-evaluate the limit. \begin{align*} \lim_{x\to \infty} \frac{% \ln\left(% \sech x \right) } { \ln x } % & = \lim_{x\to \infty} \frac{% \blue{% \frac d {dx} \left[% \ln\left(% \sech x \right) \right] } } {% \red{% \frac d {dx} \left(% \ln x \right) } } \\[6pt] % & = \lim_{x\to \infty} \frac{% \blue{% \frac 1 {% \sech x } \cdot \left(% -\sech x \tanh x \right) } } {% \red{% 1/x } } \\[6pt] % & = \lim_{x\to \infty} \frac{% \frac 1 {% \cancelred{% \sech x } } \cdot \left(% -\cancelred{% \sech x } \, \tanh x \right) } {% 1/x } \\[6pt] % & = \lim_{x\to \infty} -x\tanh x \\[6pt] % & = -\infty \cdot \tanh \infty \\[6pt] % & = -\infty \cdot (1) \\[6pt] % -\infty \end{align*} Step 5 Evaluate the original limit using the values we've found. \begin{align*} \lim_{x\to \infty} \left(% \sech x \right)^{% 1/\ln x } % & = e^{% \blue{% \lim_{x\to \infty} \ln\left(% \sech x \right)^{% 1/\ln x } } } && \text{ From Step 2 } \\[6pt] % & = e^{-\infty} && \text{ From Step 5 } \\[6pt] % & = 0 \end{align*} Answer \displaystyle \lim_{x\to \infty} \left(% \sech x \right)^{% 1/\ln x } = 0 Step 1 Evaluate the limit in its current form. \displaystyle\lim_{x\to a} \blue u^{\red v} = \blue 0^{\red{-\infty}} Step 2 Rewrite the limit using the exponential and natural log functions, and pass the limit into the exponent of e . \begin{align*} \lim_{x\to a} \blue{u^v} & = \lim_{x\to a} e^{\ln\left(\blue{u^v}\right)}\\[6pt] & = e^{\lim_{x\to a} \ln\left(\blue{u^v}\right)} \end{align*} Step 3 Focus on just the exponent of e . Simplify using the properties of logarithms and re-evaluate the limit. \begin{align*} \lim_{x\to a} \ln\left(u^v\right) & = \lim_{x\to a} v\,\ln u\\[6pt] & = \left(\lim_{x\to a} v\right)\left(\lim_{x\to a} \ln u \right)\6pt] & = -\infty \cdot \ln 0\\[6pt] & = -\infty \cdot (-\infty)\\[6pt] & = \infty \end{align*}

Step 4

Evaluate the original limit using the values we've found.

\begin{align*} \lim_{x\to a} u^v & = e^{\blue{\lim_{x\to a} \ln\left(u^v\right)}}\\[6pt] & = e^\infty\\[6pt] & = \infty \end{align*}

Limits that result in the $$0^{-\infty}$$ form will always end up being "equal" to $$\infty$$.