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I am trying to do random problems in my book and I do not know what to do for this one.

I am suppose to find the limit as x approaches infinity of $\tanh x$

I really do not know what to do I know the problem is $ \frac{(e^x - e^{-x})/2}{(e^x + e^{-x})/2}$ which I think can reduce to $ \frac{e^x - e^{-x}}{e^x + e^{-x}}$.

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2 Answers 2

up vote 2 down vote accepted

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

Where the last equality follows by multiplying by $\frac{e^{-x}}{e^{-x}}=1$. Then, since both the numerator and denominator have limit 1 as $x\rightarrow\infty$, we can conclude that $\lim_{x\rightarrow\infty} \tanh(x)=1$.

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I'll give an intuitive idea:

$\sinh(x) = \frac{e^{x}-e^{-x}}{2}$

$\cosh(x) = \frac{e^{x}+e^{-x}}{2}$

Note that for (maybe even not so) large $x$, the $e^{-x}$ terms gets extremely small. So, $\tanh(x)$ "looks like"

$\tanh(x) = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \approx \frac{e^x}{e^x}=1$

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