Prove using theorem of monotone convergence that limit below converges, in other words, prove analytically that limit below converges:
$$\lim_{x\to \infty} \frac{5^x}{2^{x^2}}$$
I tried doing: $$\lim_{x\to \infty} \frac{x \cdot \ln{5}}{x^2\cdot \ln{2}}=0$$ but I don't know if this is allowed by the theorem.
On
We could try the following:
$$\lim_{x\to\infty}\frac5{2^x}=0$$
so there's $\;M\in\Bbb R^+\;$ such that $\;x>M\implies \cfrac5{2^x}\le q<1\;$ , and thus
$$\lim_{x\to\infty\\x>M}\frac{5^x}{2^{x^2}}=\lim_{x\to\infty\\x>M}\left(\frac 5{2^x}\right)^x\le\lim_{x\to\infty\\x>M} q^x=0$$
Fill in details in the above.
Hint:
Rewrite $5^x$ as $2^{x\log_2 5}$, so that $$\frac{5^x}{2^{x^2}}=2^{x\log_25-x^2}$$ (You don't really need the monotone convergence theorem, only the continuity ot the function $t\mapsto 2^t$.)
Some more details:
As a polynomial is asymptotically equivalent to its leading term, $\;\lim_{x\to +\infty}(x\log_25-x^2)=\lim_{x\to +\infty}(-x^2)=-\infty$, so $$\lim_{x\to +\infty}\frac{5^x}{2^{x^2}}=\lim_{x\to +\infty}2^{x\log_25-x^2}=\lim_{u\to -\infty}2^u=0.$$