MHB Calculating Hyperbolic Limit of $\frac{x}{\cosh{x}}$

[Guest2]

How do you calculate the limit $\displaystyle \lim_{x \to \infty}\frac{x}{\cosh{x}}$

 

[Greg]

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

 

[Guest2]

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

Thanks. Is this correct?

$\displaystyle \lim_{x \to \infty}\frac{x}{\cosh{x}} = \frac{1}{2} \lim_{x \to \infty} \frac{xe^{-x}}{1+e^{-2x}} = \frac{0}{1+0} = 0.$

 

[Prove It]

How do you calculate the limit $\displaystyle \lim_{x \to \infty}\frac{x}{\cosh{x}}$

This is an $\displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$ indeterminate form, so you can use L'Hospital's Rule...

$\displaystyle \begin{align*} \lim_{x \to \infty} \frac{x}{\cosh{(x)}} &= \lim_{x \to \infty} \frac{\frac{\mathrm{d}}{\mathrm{d}x} \, \left( x \right) }{\frac{\mathrm{d}}{\mathrm{d}x}\,\left[ \cosh{(x)} \right] } \textrm{ by L'Hospital's Rule} \\ &= \lim_{x \to \infty} \frac{1}{\sinh{(x)}} \\ &= 0 \end{align*}$