A lot of terms to calculate lim I'm trying to prepare for exam and I came across a limit to calculate. 
$$
\lim_{n->\infty} \frac{2^n + \left(1+\dfrac{1}{n^2}\right)^{n^3} + \dfrac {4^n}{n^4}}{\dfrac {4^n}{n^4} + n^3\cdot 3^n}
$$
When I'm trying to extract $4^n$ I end up with nothing. 
And I managed to tell that $(1+\frac{1}{n^2})^{n^3}$ goes to infinity, bounding it with $2$ and $3$. Cause I know that $(1+\frac{1}{n^2})^{n^2}$ goes to $e$ as $n$ is increasing. 
I'd appreciate some help or tips on that one, cause I'm stuck with it for over an hour and I couldn't find any clues how to solve it on-line.
Chris

$$\frac{2^n + e^n + \frac{4^n}{n^4}}{\frac{4^n}{n^4} + 3^nn^3} = \frac{2^nn^4 + e^nn^4 + 4^n}{4^n + n^73^n} = \frac{\frac{n^4}{2^n} + (\frac{e}{4})^n\cdot n^4 + 1}{1 + (\frac{3}{4})^nn^7} $$
Now, since $a^n$ is an infinite of magnitude greater than $n^b$ for all $b$, we conclude that all those fractions tends to zero (cause also $a^n$ tends to zero really fast if $a < 1$) that leaves you with $1$ both at the numerator and the denominator
So the limit is $1$
