How do we show that the following limit goes to infinity?

Question

I am reading a certain proof and along the way, I encountered the following

$$\lim_{x \to \infty} \frac{e^{-\sqrt{x}}\sinh(cx)}{x} = +\infty$$

for any $c>0$.

The author did not explain why this limit is true and I cannot see the reason why this is true. Can someone explain to me why?

Answer 1

In fact, the limit depends on the sign of $c$.

Suppose $c >0$; when $x$ is large $$\sinh(cx)\approx \frac 12 e^{cx}\implies \frac{e^{-\sqrt{x}}\sinh(cx)}{x}\approx\frac{e^{c x-\sqrt{x}}}{2 x}$$ Consider then $$f(x)=\frac{e^{c x-\sqrt{x}}}{2 x}$$ $$f'(x)=\frac{e^{c x-\sqrt{x}} \left(2 c x-\sqrt{x}-2\right)}{4 x^2}$$ The first derivative cancels at $$x_{1,2}=\frac{8 c+1\pm\sqrt{16 c+1}}{8 c^2}$$ and, for $x>x_2$, the first derivative is always positive.

Taking logarithms, $$\log(f(x))=c x- \sqrt x-\log(x)-\log(2)$$ shows the same.

Answer 2

Using $\sinh cx=\frac{e^{cx}-e^{-cx}}{2}$ $$\lim_{x \to \infty} \frac{e^{-\sqrt{x}}\sinh(cx)}{x} =\lim_{x \to \infty}\frac{e^{cx-\sqrt{x}}}{2x}-\lim_{x \to \infty}\frac{e^{-cx-\sqrt{x}}}{2x}=+\infty$$ Since $cx-\sqrt{x}>0$ for large $x$.

Answer 3

If you just want to convince yourself it's true without formally proving it, note that for $c>0$ and large $x$, $\sinh{cx} \approx \frac{e^{cx}}{2}$, so you have something that grows like $(e^{cx-\sqrt{x}})/2x$. But for any fixed value of $c$, the $cx$ term will dominate $-\sqrt x$ when $x$ is large, i.e. you are looking at $e^{cx}/2x$ when $x$ is large. So the numerator grows exponentially, which is faster than the denominator (which is linear).