# Anakin1369's question at Yahoo! Answers regarding a limit having an indeterminate form

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#### MarkFL

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Re: Anakin1369's question at Yahoo! Answers regardina limit having indeterminate form

Hello Anakin1369,

We are given to evaluate:

$$\displaystyle \lim_{x\to\infty}\tanh^x(x)=L$$

Observing that we have the indeterminate form $$\displaystyle 1^{\infty}$$, I recommend taking the natural log of both sides:

$$\displaystyle \ln\left(\lim_{x\to\infty}\tanh^x(x) \right)=\ln(L)$$

Since the natural log function is continuous, we may "bring it inside the limit" to get:

$$\displaystyle \lim_{x\to\infty}\ln\left(\tanh^x(x) \right)=\ln(L)$$

Now, using the log property $$\displaystyle \log_a\left(b^c \right)=c\cdot\log_a(b)$$ we may write:

$$\displaystyle \lim_{x\to\infty}x\ln\left(\tanh(x) \right)=\ln(L)$$

Bringing the $x$ out front down into the denominator, we have:

$$\displaystyle \lim_{x\to\infty}\frac{\ln\left(\tanh(x) \right)}{\frac{1}{x}}=\ln(L)$$

Now we have the indeterminate form $$\displaystyle \frac{0}{0}$$, and so application of L'Hôpital's rules gives:

$$\displaystyle \lim_{x\to\infty}\frac{\text{csch}(x)\text{sech}(x)}{-\frac{1}{x^2}}=\ln(L)$$

$$\displaystyle -\lim_{x\to\infty}\frac{x^2}{\sinh(x)\cosh(x)}=\ln(L)$$

The exponential function in the denominator "dominates" the quadratic in the numerator, hence we have:

$$\displaystyle 0=\ln(L)$$

Converting from logarithmic to exponential form, we then find:

$$\displaystyle L=e^0=1$$

and so we may conclude:

$$\displaystyle \lim_{x\to\infty}\tanh^x(x)=1$$

To Anakin1369 and any other guests viewing this topic, I invite and encourage you to post other calculus problems in our Calculus forum.

Best Regards,

Mark.

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