How to solve lim as x approaches infinity for $[\tanh(x)]^x$

Question

I got as far as lim x approaches infinity for $\ln y = x \ln(\tanh x)$. I'm not sure what to do there. I know $\tanh x$ as $x$ approaches infinity is one but $1^\infty$ isn't the correct answer.

So what I did was I took the $lim x→∞ lny = lim x→∞ x lm \tanh x$. I know that the $$\tanh x=(e^x-1)/(e^x+1)$$ but plugging that in, I get $$\lim x→∞ (x\ln (∞/∞)) $$

I am not sure what I am supposed to do there or if I even did it right. We're not allowed to take the logs or exp(...) because we haven't learned that yet. I am just wondering if there is a simple way to set up the problem?

I know what the tanhx graph looks like and the limit approaches one. But is that enough to suffice as evidence to say that $(\tan x)^x$ have to approach one? Because just plugging it in, I would get $1^∞$ and I know that is indeterminate.

Answer 1

Use the fact that $$ \tanh x=\frac{e^{2x}-1}{e^{2x}+1}=1-\frac{2}{e^{2x}+1} $$ Now do the substitution $t=e^{2x}+1$, so $2x=\log(t-1)$ and your limit becomes $$ \lim_{t\to\infty}\sqrt{\left(1-\frac{2}{t}\right)^{\!\log(t-1)}}= \lim_{t\to\infty}\sqrt{ \left(\left(1-\frac{2}{t}\right)^{\!t}\right)^{\!\tfrac{\log(t-1)}{t}}} $$ This is not an indeterminate form any more, because $$ \lim_{t\to\infty}\left(1-\frac{2}{t}\right)^{\!t}=e^{-2}, \qquad \lim_{t\to\infty}\frac{\log(t-1)}{t}=0 $$

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With l'Hôpital, compute the limit of the logarithm, that is, $$ \lim_{x\to\infty}x\log\tanh x= \lim_{x\to\infty}\frac{\log\tanh x}{1/x} $$ This is of the form $0/0$, because $\lim_{x\to\infty}\tanh x=1$, so we can write it as $$ \lim_{x\to\infty}\frac{\log\sinh x-\log\cosh x}{1/x} \overset{\mathrm{(H)}}{=} \lim_{x\to\infty}\frac{\dfrac{\cosh x}{\sinh x}-\dfrac{\sinh x}{\cosh x}}{-1/x^2} $$ Now simplify the numerator: $$ \dfrac{\cosh x}{\sinh x}-\dfrac{\sinh x}{\cosh x}= \frac{\cosh^2x-\sinh^2x}{\sinh x\cosh x}=\frac{1}{\sinh x\cosh x} $$ so we can go on $$ =\lim_{x\to\infty}\frac{-x^2}{\sinh x\cosh x}= \lim_{x\to\infty}\frac{-2x^2}{\sinh2x} \overset{\mathrm{(H)}}{=} \lim_{x\to\infty}\frac{-4x}{2\cosh2x} \overset{\mathrm{(H)}}{=} \lim_{x\to\infty}\frac{-4}{4\sinh2x}=0 $$ Since $\lim_{x\to\infty}\log\bigl((\tanh x)^x\bigr)=0$, we conclude $$ \lim_{x\to\infty}(\tanh x)^x=e^0=1 $$

Answer 2

Just simple calculation:

$\because \tanh{x} = \frac{e^{2x}-1}{e^{2x}+1}$

$\therefore \lim \limits_{x \to \infty} (\tanh{x})^x$

$=\lim \limits_{x \to \infty} (\frac{e^{2x}-1}{e^{2x}+1})^x$

$=\lim \limits_{x \to \infty} (1-\frac{2}{e^{2x}+1})^x$

$= \lim \limits_{x \to \infty} [(1-\frac{2}{e^{2x}+1})^\frac{e^{2x}+1}{-2}]^\frac{-2x}{e^{2x}+1}$

$= \lim \limits_{x \to \infty} e^\frac{-2x}{e^{2x}+1}$, $\because e^x$ is continuous

$=e^0$