Limit of a Cosh function Evaluate
$$\lim_{t\to\infty} (\cosh x)^{1/x}.$$ I tried to use L'Hopital's but I think I made a mess of the differentiation, and the differentiation doesn't seem like it'll help much.
 A: Note that for positive $x$ we have 
$$\cosh x=\frac{e^x+e^{-x}}{2}\lt \frac{e^x+e^x}{2}=e^x.$$
Also, 
$$\cosh x\gt \frac{e^x}{2}.$$
Thus
$$(1/2)^{1/x}e \lt (\cosh x)^{1/x}\lt e.$$
Since $\lim_{x\to\infty}(1/2)^{1/x}=1$, by squeezing the desired limit is $e$.
Added: In a comment, OP has asked whether one can solve this using L'Hospital's Rule. so here goes. 
Note that $\ln((\cosh x)^{1/x})=\frac{\ln(\cosh x)}{x}$. The numerator an denominator both blow up, and are differentiable, so we can use L'Hospital's Rule. We get
$$\lim_{x\to\infty}\frac{\ln(\cosh x)}{x}=\lim_{x\to\infty}\frac{\frac{\sinh x}{\cosh x}}{1}.$$
For the limit on the right, after some cancellation we can rewrite $\\frac{\sinh x}{\cosh x}$ as
$$\frac{e^x-e^{-x}}{e^x+e^{-x}}.$$
Divide top and bottom by $e^x$. So we want
$$\lim_{x\to\infty}\frac{1-e^{-2x}}{1+e^{-2x}}.$$
The limit is $1$. 
Since the $\ln$ function is continuous, and $\lim_{x\to\infty}\ln((\cosh x)^{1/x})=1$, it follows that $\lim_{x\to\infty} (\cosh x)^{1/x}=e^1=e$. 
