Evaluating $ \lim_{n\to\infty} \frac{5^{\sqrt{n}}}{2^n} $ 
Consider the limit
  $$
\lim_{n\to\infty} \frac{5^{\sqrt{n}}}{2^n}
$$
  How to find it? 

My attempt was to take a logarithm from this expression and get
$$
e^{\lim_{n\to\infty}\ln\left({\frac{5^{\sqrt{n}}}{2^n}}\right)}
$$
Then I used rules of logarithms and concluded that the limit in the exponent is $-\infty$. Can I conclude from here that the actual limit is equal to $0$?
 A: Yes, you can conclude the limit is $0$; the limit is clearly positive allowing you to write it in exponential form.
Alternatively, note that $5<2^3$, which implies $\frac{5^{\sqrt n}}{2^n} < \frac{2^{3 \sqrt n}}{2^n} = 2^{3\sqrt n - n}$. Noting that $3\sqrt n - n$ approaches $-\infty$, we conclude the limit is $0$.
A: Yup, that's perfectly fine. We can rewrite the function in the limit as the exponential of its natural logarithm, and then use properties of the logarithm to rewrite it. The continuity of the exponential lets us bring the limit inside the exponential, and thus
$$\begin{align}
\lim_{n \to \infty} \frac{5^\sqrt n}{2^n} &= \lim_{n \to \infty} \exp\left(  \ln \left( \frac{5^\sqrt n}{2^n}\right) \right) \\
&= \lim_{n \to \infty} \exp\left( \sqrt n \ln(5) - n \ln(2)\right) \\
&=  \exp\left( \lim_{n \to \infty} \sqrt n \ln(5) - n \ln(2)\right)
\end{align}$$
Intuitively, $n$'s term dominates the growth of this new limit, so it should be $-\infty$. Asymptotics allows us to formally argue this since
$$\lim_{n \to \infty} \frac{\sqrt n \ln(5) - n \ln(2)}{-n \ln(2)} = \lim_{n \to \infty} \frac{\ln(5)}{-\sqrt n \ln(2)}+1=1$$
establishing that the numerator and denominator of the first expression are asymptotically equivalent and thus have the same infinite limit (which is significantly more obvious for the bottom function).
Thus, the limit in the exponential approaches $-\infty$ and in turn the function overall has its limit approach zero, i.e.
$$\lim_{n \to \infty} \frac{5^\sqrt n}{2^n} = \exp\left( \lim_{n \to \infty} \sqrt n \ln(5) - n \ln(2)\right) = \lim_{n\to -\infty} e^n = 0$$
It feels like there is a simpler approach to this that we're overlooking though, but in any case - yes, your method is fine.
A: A possible way to verify the limit without logarithms is as follows using the fact


*

*$(\star):\frac{n}{3}\geq \sqrt{n} \Leftrightarrow n \geq 9$
It follows for $n \geq 9$
$$0 \leq \frac{5^{\sqrt{n}}}{2^n}= \frac{5^{\sqrt{n}}}{8^{\frac{n}{3}}} \stackrel{(\star)}{\leq} \frac{5^{\sqrt{n}}}{8^{\sqrt{n}}}= \left(\frac{5}{8} \right)^{\sqrt{n}}\stackrel{n \to \infty}{\longrightarrow}0$$
