# Show that $e^{\varepsilon |x|^{\varepsilon}}$ grows faster than $\sum_{k=0}^{\infty} {|x|^{2k}}/{(k!)^2}$

I am wondering whether we have for $$f(x):=\sum_{k=0}^{\infty} \frac{|x|^{2k}}{(k!)^2}$$ that

$$\lim_{x \rightarrow \infty} \frac{e^{\varepsilon |x|^{\varepsilon}}}{f(x)} = \infty$$ for any $\varepsilon>0$?

I assume that this is true as factorials should somehow outgrow powers, but I do not see how to show this rigorously?

Does anybody have an idea?

You have that, for $x>0$, $$\sum_{k=0}^\infty \frac{x^{2k}}{(k!)^2} \ge\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}=\cosh(x)$$ and $\cosh x \sim_{\infty} e^x$. So it doesn't work for $\varepsilon <1$