# Calculate $\lim\limits_{x \to \infty} x-\log(\cosh(x))$ for

I need help with calculating the following limit: $\lim\limits_{x \to \infty} x-\log(\cosh(x))$

So far, I've figures that $\lim\limits_{x \to \infty}x-\log(\cosh(x))=\lim\limits_{x \to \infty}(x-\log(e^x-e^{-x}))+\lim\limits_{x \to \infty}\log(2)$

I've tried to rewrite algebraically back and forth, but I'm stuck in trying to reduce it to something I recognize the limits of. A hint would be much appreciated.

• How is n and x conntected? – Yuriy S Feb 19 '16 at 20:46
• Sorry, it was a typo, I meant of course x instead of n – jta Feb 19 '16 at 20:48

## 3 Answers

Hint: $$x - \log\left(e^x - e^{-x}\right) = \log\left(\frac{e^x}{e^x - e^{-x}}\right) = \log\left(\frac{1}{1 - e^{-2x}}\right)$$

Hint: write as$$\lim\limits_{x\to \infty}\log(\exp(x-\log(\cos h x)))=\lim\limits_{x\to \infty}\log \bigg(\frac{e^x}{\cos h x}\bigg)=\dots$$

You were almost there

\begin{array}{rcl} \ln(2) + \lim_{x \to \infty} \ x-\ln(e^x-e^{-x}) &=& \ln(2) + \lim_{x\to \infty} \ x-\ln(\ e^x\cdot(1-e^{-2x})\ ) \\ &=& \ln(2) + \lim_{x\to \infty} x-x + \ln(1-e^{-2x}) \\ &=& \ln(2) \end{array}