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:

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

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

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

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

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

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

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

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

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

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

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

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

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

$$0=\ln(L)$$

Converting from logarithmic to exponential form, we then find:

$$L=e^0=1$$

and so we may conclude:

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

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.