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Find $\lim\limits_{n\rightarrow \infty} \frac{n^t}{e^n}$ where $t$ is a positive real number.

My answer is that the expression is indeterminate ($\frac{\infty}{\infty}$) but it will tend to $0$ because as $n$ gets large, the exponential term will dominate the polynonial term in the expression, so the expression will tend to $0$ as $n$ gets large. This answer is also the same as the solution provided in my text.

But I am wondering what is the actual workings to solve this? I'd know to because I want to be familiar with all the tricks to solve limits where the expression is indeterminate.

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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.

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Hint

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

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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:

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

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

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Expand $e^n$ and then divide both numerator and denominator by $n^t$. The expansion of $e^n/n^t$ will tend to $\infty$ when $n$ goes to $\infty$.

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There is a simple trick: exploit the fact that $e^x\geq 1+x$. We have: $$0\leq\frac{n^t}{e^n}=\frac{n^t}{\left(e^{\frac{n}{t+1}}\right)^{t+1}}\leq \frac{n^t}{\left(1+\frac{n}{t+1}\right)^{t+1}}\leq\frac{(t+1)^{t+1}}{n},$$ so the limit is zero by squeezing.

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