Anakin1369's Limit of (tanh(x))^x: Yahoo Answers

[MarkFL]

Here is the question:

What is the limit of (tanh(x))^x as x approaches infinity?

Hi. If you could provide me with the process that leads to the answer that would really help. Thanks.

Here is a link to the question:

What is the limit of (tanh(x))^x as x approaches infinity? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

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 http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.