How to solve $\lim \left(\frac{n^3+n+4}{n^3+2n^2}\right)^{n^2}$ I can't seem to find a way to solve:
$$\lim \left(\dfrac{n^3+n+4}{n^3+2n^2}\right)^{n^2}$$
I've tried applying an exponential and logaritmic to take the $n^2$ out of the exponent, I've tried dividing the expression, but I don't get anywhere that brings light to the solution.
Any ideas?
 A: $$\lim_{n \rightarrow \infty} \log f(n)  = n^2 \log\dfrac{n^3+n+4}{n^3+2n^2}= n^2 \log (1 - O(\frac{1}{n})) \rightarrow -\infty $$
Hence $$\lim_{n \rightarrow \infty} f(n) = e^{-\infty} = 0$$
A: $$
\begin{align*}
L &= \lim_{n\to\infty}\left(\frac{n^3+n+4}{n^3+2n^2}\right)^{n^2} \\ 
&= \lim_{n\to\infty}\left(\frac{n^3 +2n^2 - 2n^2+n+4}{n^3+2n^2}\right)^{n^2}\tag1 \\
&= \lim_{n\to\infty}\left(1 + \frac{- 2n^2+n+4}{n^3+2n^2}\right)^{n^2} \tag2 \\
&= \lim_{n\to\infty}\left(1 + \frac{- 2n^2+n+4}{n^3+2n^2}\right)^{\frac{n^2(n^3+2n^2)(n-2n^2+4)}{(n^3+2n^2)(n-2n^2+4)}} \tag3 \\
&= \lim_{n\to\infty}e^{n^2(n-2n^2+4)\over(n^3+2n^2)} \tag4
\end{align*}
$$
Now consider:
$$
\lim_{n\to\infty}{n^2(n-2n^2+4)\over(n^3+2n^2)} = -\infty
$$
Hence your limit is:
$$
e^{-\infty} = 0
$$
Description of steps:


*

*$(1)$ add and subtract $2n^2$

*$(2)$ perform division

*$(3)$ multiply the power by the reciprocal of the fraction inside parentheses

*$(4)$ use the limit for $(1 + {1\over x^n})^{x_n}$ when $x_n \to\ \infty$
A: I suppose that $n\to \infty$
$$\lim_{n\to \infty} \left(\dfrac{n^3+n+4}{n^3+2n^2}\right)^{n^2}=\lim_{n\to \infty} \Biggl(\left(1+\dfrac{-2n^2+n+4}{n^3+2n^2}\right)^{\dfrac{n^3+2n^2}{-2n^2+n+4}}\Biggr)^{\frac{-2n^4+n^3+4n^2}{n^3+2n^2}}=\ e^{-\infty}=0$$
A: We have $$ \lim_{n\to\infty}\left(\frac{n^3+n+4}{n^3+2n^2}\right)^{n^2} =
 \lim_{n\to\infty}\left(\frac{1+\frac{n+4}{n^3} }{1+\frac{2n^2}{n^3}}\right)^{n^2}=
\frac{e}{\lim_{n\to\infty}e^n}$$
Hence your limit is: 0
