Test $x_n=\frac{n^2 e^{-\sqrt{n}}}{\cos(1/n)}$ for convergence and give its limit if possible.

Question

Test $x_n=\frac{n^2 e^{-\sqrt{n}}}{\cos(1/n)}$ for convergence and give its limit if possible.

Easy part first: $1/n\rightarrow 0$ as $n\rightarrow\infty$ and hence $\cos(1/n)\rightarrow\cos(0)=1$.

Now the "harder" part. Would you say this arugment is sufficient/correct?

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We rewrite the numerator as $n^2 \frac{1}{e^\sqrt n}$. Since $e^\sqrt n$ grows faster than $n^2$ the term converges towards $0$ for $n\rightarrow\infty$.

So we have $lim_{n\rightarrow\infty}x_n=0/1=0$.

Answer 1

With $\sqrt{n}=t$, you can write the term $n^2 \frac{1}{e^\sqrt n}$ as $\frac{t^4}{e^t}$ and when $t$ goes to infinity, you may apply L'Hospital Rule 4 times to see that this limit is zero. Based upon that observation you are right to say that a polynomial terms "loses" against an e-power. This answers the "grows faster" part of your question.

Answer 2

If we take logarithm of the numerator, we get

$$-\sqrt{n}+2\ln (n)=-\sqrt {n}+4\ln (\sqrt {n}) $$ $$=\sqrt {n}(4\frac {\ln (\sqrt {n})}{\sqrt {n}}-1) $$

which goes to $-\infty $.

the limit is zero.