<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] This is right, because it depends what is in your base. Consider
$ (e^2)^{\ln(n)} = (e^{\ln(n)})^2 = n^2 $
Since $10 > e^2$ or rather $\ln(10) = a > 2$
You have: $10^{\ln(n)}$ is equivalent of $ n^a $ and therefore
$ \frac { 10^{\ln(n)}}{n^2+1}   \rightarrow +\infty$
In your initial example $2 < e < e^2$ so the limit is zero.
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