What is the workings to solve $\lim\limits_{n\rightarrow \infty} \frac{n^t}{e^n}$?

Solution 1:

Consider $$A=\frac{n^t}{e^n}$$ Taking the logarithms of both sides $$B=\log(A)=t \log(n)-n$$ Function $B$ has a derivative equal to $$B'=\frac{t}{n}-1$$ which is negative as soon as $n>t$; so, the function decreases and $B$ goes to $-\infty$ ans so $A$ goes to $0$.

Using derivative could have been done with function $A$; $$A'=e^{-n} n^{t-1} (t-n)$$ which cancels if $t=n$ and, at this point, the value of the second derivative is $-e^{-t} t^{t-1}$ so it was a maximum.

Solution 2:

Hint

Apply the L'Hôpital's rule $\lfloor t\rfloor$+1 times and the indeterminate form disappears and we get the limit is $0$.

Solution 3:

Take the positive infinite series

$$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{n^t}{e^n}\;,\;\;\;t>0$$

and apply the quotient test to check whether it converges:

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^t}{e^{n+1}}\frac{e^n}{n^t}=\left(1+\frac1n\right)^t\frac1e\xrightarrow[n\to\infty]{}\frac1e<1\implies$$

$$ \implies \;\;\;\sum_{n=1}^\infty a_n\;\;\; \text{converges}\;\;\implies\;\;a_n\xrightarrow[n\to\infty]{}0$$