# 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?

• Hint: put $n=1/t$ where $t>0$ is small. – user1892304 Jan 25 at 11:56
• No \dfrac in titles please. – Did Jan 25 at 12:32
• limits at which point? you didn't mention the point.... :( – Abhas Kumar Sinha Jan 25 at 15:25
• @AbhasKumarSinha, when using $n$ we usually assume $n \to \infty$ so I omitted that part. – Concept7 Jan 25 at 17:32

\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 $$x_n \to\ \infty$$
• Some extra work need to be taken care of in step 4, the limit is $(1+\frac{1}{x^n})^{x_n y_n}\to e^{y_n}$ when $x_n\to \infty$, this is not trivial and I'm not sure it is always true. – P. Quinton Jan 25 at 12:13
• @P.Quinton you may justify this by continuity of $a^x$. Hence $\lim a^{x_n} = a^{\lim x_n}$ – roman Jan 25 at 12:13
• I was actually talking about the step just before that, the fourth – P. Quinton Jan 25 at 12:14
• @P.Quinton well that would be a good candidate for the OP to consider proving it – roman Jan 25 at 12:24

$$\lim_{n \rightarrow \infty} \log f(n) = n^2 \log\dfrac{n^3+n+4}{n^3+2n^2}= n^2 \log (1 - O(\frac{1}{n})) \rightarrow -\infty$$ Hence $$\lim_{n \rightarrow \infty} f(n) = e^{-\infty} = 0$$

I suppose that $$n\to \infty$$

$$\lim_{n\to \infty} \left(\dfrac{n^3+n+4}{n^3+2n^2}\right)^{n^2}=\lim_{n\to \infty} \Biggl(\left(1+\dfrac{-2n^2+n+4}{n^3+2n^2}\right)^{\dfrac{n^3+2n^2}{-2n^2+n+4}}\Biggr)^{\frac{-2n^4+n^3+4n^2}{n^3+2n^2}}=\ e^{-\infty}=0$$

We have $$\lim_{n\to\infty}\left(\frac{n^3+n+4}{n^3+2n^2}\right)^{n^2} = \lim_{n\to\infty}\left(\frac{1+\frac{n+4}{n^3} }{1+\frac{2n^2}{n^3}}\right)^{n^2}= \frac{e}{\lim_{n\to\infty}e^n}$$