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$$

\lim_{x\to\infty} tanh(x)=\lim_{x\to\infty} sinh(x)/cosh(x)=\lim_{x\to\infty} \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}

$$

Can you just assume that the [tex]e^{-x}[/tex] will approach 0 and the [tex]e^{x}[/tex] will cancel and the limit will reduce to 1?

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# Homework Help: Limit problem

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