<s>[INST] Limit of $\frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$ as $n\to\infty$ I've tried to solve the limit
$$     \lim_{n \to \infty} \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$$
but I'm not sure.
$$ \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1} = \frac { 2^{\sqrt{ (\ln n)^2+ 2\ln n}}}{n^2+1} =  \frac { 2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}}}{n^2+1} \sim  \frac { 2^{\ln n }}{n^2+1}   \rightarrow 0$$
Is it right?
I have another exercize that ends similarly with 
$$ \frac { 10^{\ln n }}{n^2+1}   \rightarrow 0$$
But the book says that the result is $+\infty$.
 [/INST] Your intuition is correct, but the part where you employ the symbol $\sim$ lacks a rigorous justification. To fully go into detail, you have the following expressions and estimates:
$$\frac{2^{\ln{n}\sqrt{1+\frac{2}{\ln{n}}}}}{n^2+1}=\frac{n^{{\ln{2}} \sqrt{1+\frac{2}{\ln{n}}}}}{n^2+1} \leqslant \frac{n^{\sqrt{3}\ln{2}}}{n^2+1}$$ for any $n \geqslant 3$, easily obtained by majorizing the exponent of $n$ in the numerator.
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