Showing $\lim_{x \to \infty}\left(\;\cosh \sqrt{x+1}-\cosh{\sqrt{x}}\;\right)^{1/\sqrt{x}}=e$ Trying to compute
$$\lim_{x \to \infty}\left(\;\cosh \sqrt{x+1}-\cosh{\sqrt{x}}\;\right)^{1/\sqrt{x}}=e$$
I arrived to the equivalent expression $$\lim_{x \to \infty}\frac{1}{\sqrt{x}}\log\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}}   \right)$$but couldn't undo the indeterminate form $0 \cdot \infty$. L'Hôpital seems to become a loop.
Any suggestions?
Thanks in advance.
 A: Let's take a closer look at
\begin{align}
2(\cosh(\sqrt{x+1})-\cosh(\sqrt{x}))&=
e^{\sqrt{x+1}}+e^{-\sqrt{x+1}}
-e^{\sqrt{x}}-e^{-\sqrt{x}}\\
&\sim_{x \infty}
e^{\sqrt{x+1}}
-
e^{\sqrt{x}}\\
&\sim_{x \infty}
e^{\sqrt{x}}(-1+e^{\sqrt{x+1}-\sqrt{x}})\\
&\sim_{x \infty}
e^{\sqrt{x}}(-1+e^{\sqrt{x}(\sqrt{1+1/x}-1)})\\
&\sim_{x \infty}
e^{\sqrt{x}}(-1+e^{1/(2\sqrt{x})})\\
&\sim_{x \infty} \frac{e^{\sqrt{x}}}{2\sqrt{x}}.\\
\end{align}
Since $\frac{-1}{\sqrt{x}}\log(4\sqrt{x}) \to 0$, we have
$$
\frac{1}{\sqrt{x}}\log\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}}   \right)
\sim \frac{1}{\sqrt{x}}\sqrt{x}=1,
$$
and by applying $e^\cdot$
$$
\lim_{x \to \infty}\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}}   \right)^{\frac{1}{\sqrt{x}}}=e.
$$
A: $$\lim_{x \to \infty}\frac{1}{\sqrt{x}}\log\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}}   \right) = \lim_{x \to \infty} \frac{1}{\sqrt{x}}\log\left\{\frac{1}2\left(e^{\sqrt{x+1}}+e^{-\sqrt{x+1}}+e^{\sqrt{x}}+e^{-\sqrt{x+1}}\right)\right\}$$
It's easy to see that the contribution of the addition of the exponentials will be insignificant (for instance, divide by the limit without them and apply L'Hop to get $1$, or simply bound them by $1$ and use inequalities to see the growth is neglible when divided by $\sqrt{x}$. Hence we write this
$$\lim_{x \to \infty} \frac{1}{\sqrt{x}}\log\left\{\frac{1}2\left(e^{\sqrt{x+1}}+e^{\sqrt{x}}\right)\right\}$$
Again, comparing this to the same expression with the ${x+1}$ replaced with an $x$, it's easy to bound the result as say a constant time the exact argument of the logarithm, which will be negligible when divided out by the $\sqrt{x}$. So this limit is just
$$ \lim_{x \to \infty} \frac{1}{\sqrt{x}}\log\left\{\frac{1}2\left(e^{\sqrt{x}}+e^{\sqrt{x}}\right)\right\} = \lim_{x\to\infty} \frac{1}{\sqrt{x}}\log\left\{e^{\sqrt{x}}\right\} = 1.  $$
Which readily yields the desired result
$$ \lim_{x \to \infty}\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}}   \right)^{\frac{1}{\sqrt{x}}}=e$$
A: $$A=\cosh \left(\sqrt{x+1}\right)-\cosh \left(\sqrt{x}\right)=2 \sinh \left(\frac{\sqrt{x+1}-\sqrt{x}}{2} \right) \sinh
   \left(\frac{\sqrt{x}+\sqrt{x+1}}{2} \right)$$ When $x$ is large, using equivalents,
$$\sinh \left(\frac{\sqrt{x+1}-\sqrt{x}}{2} \right)\sim\frac 1{4 \sqrt x}$$
$$\sinh\left(\frac{\sqrt{x}+\sqrt{x+1}}{2} \right)\sim \sinh(\sqrt{x})\sim \frac 12e^{\sqrt x}$$
$$A\sim 2 \times \frac 1{4 \sqrt x}\times \frac 12e^{\sqrt x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x}}$$
$$B=A^{\frac 1{\sqrt x}}\sim \Bigg[\frac{e^{\sqrt{x}}}{4 \sqrt{x}} \Bigg]^{\frac 1{\sqrt x}}$$ Take logarithms
$$\log(B)=\frac 1{\sqrt x}\log\Bigg[\frac{e^{\sqrt{x}}}{4 \sqrt{x}} \Bigg]=1-\frac 1{\sqrt x}\log(4\sqrt x)$$ and $B=e^{\log(B)}$
