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 $+\infty$.
 A: 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.
A: Your use of $\sim$ is not correct although this small mistake does not change the limit. You have
$$2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}} \stackrel{n\to\infty}{\sim}\color{blue}{2\cdot}2^{\ln n}$$
This is so because
$$\frac{2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}}}{2^{\ln n}}= 2^{\ln n\left(\sqrt{ 1+ \frac {2}{\ln n}}-1\right)} = 2^{\ln n \frac{\frac{2}{\ln n}}{\sqrt{ 1+ \frac {2}{\ln n}}+1}}$$ $$=2^{\frac{2}{\sqrt{ 1+ \frac {2}{\ln n}}+1}}\stackrel{n\to\infty}{\longrightarrow}2^1=2$$
At the end you may add another step (which is also useful for your second question):


*

*$2^{\ln n} = e^{\ln 2\cdot \ln n} = n^{\ln 2}$ and similarly for your second question $10^{\ln n} = n^{\ln 10}$
Putting this together, you get
$$0\leq \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1} \ldots \sim  \frac { 2\cdot 2^{\ln n }}{n^2+1}=\frac{2n^{\ln 2}}{n^2+1}\stackrel{\color{green}{\ln 2 < 1}}{\leq}\frac{2n}{n^2+1}\stackrel{n\to\infty}{\longrightarrow}0$$
Similarly for your second question
$$\frac { 10^{\ln n }}{n^2+1} = \frac{n^{\ln 10}}{n^2+1}\stackrel{\color{green}{\ln 10 > \frac 94}}{\geq}\frac{n^{\frac 94}}{n^2+1}$$ $$\geq\frac{1}{2}\cdot\frac{n^{\frac 94}}{n^2}=n^{\frac 14}\stackrel{n\to\infty}{\longrightarrow}+\infty$$
A: I believe you can apply L'hopitals rule since you have a situation in which there is infinity over infinity in both problems.
A: 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.
