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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?

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  • $\begingroup$ Hint: put $n=1/t$ where $t>0$ is small. $\endgroup$ – user1892304 Jan 25 at 11:56
  • $\begingroup$ No \dfrac in titles please. $\endgroup$ – Did Jan 25 at 12:32
  • $\begingroup$ limits at which point? you didn't mention the point.... :( $\endgroup$ – Abhas Kumar Sinha Jan 25 at 15:25
  • $\begingroup$ @AbhasKumarSinha, when using $n$ we usually assume $n \to \infty$ so I omitted that part. $\endgroup$ – Concept7 Jan 25 at 17:32
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$$ \begin{align*} L &= \lim_{n\to\infty}\left(\frac{n^3+n+4}{n^3+2n^2}\right)^{n^2} \\ &= \lim_{n\to\infty}\left(\frac{n^3 +2n^2 - 2n^2+n+4}{n^3+2n^2}\right)^{n^2}\tag1 \\ &= \lim_{n\to\infty}\left(1 + \frac{- 2n^2+n+4}{n^3+2n^2}\right)^{n^2} \tag2 \\ &= \lim_{n\to\infty}\left(1 + \frac{- 2n^2+n+4}{n^3+2n^2}\right)^{\frac{n^2(n^3+2n^2)(n-2n^2+4)}{(n^3+2n^2)(n-2n^2+4)}} \tag3 \\ &= \lim_{n\to\infty}e^{n^2(n-2n^2+4)\over(n^3+2n^2)} \tag4 \end{align*} $$ 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 $x_n \to\ \infty$
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  • $\begingroup$ 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. $\endgroup$ – P. Quinton Jan 25 at 12:13
  • $\begingroup$ @P.Quinton you may justify this by continuity of $a^x$. Hence $\lim a^{x_n} = a^{\lim x_n}$ $\endgroup$ – roman Jan 25 at 12:13
  • $\begingroup$ I was actually talking about the step just before that, the fourth $\endgroup$ – P. Quinton Jan 25 at 12:14
  • $\begingroup$ @P.Quinton well that would be a good candidate for the OP to consider proving it $\endgroup$ – roman Jan 25 at 12:24
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$$\lim_{n \rightarrow \infty} \log f(n) = n^2 \log\dfrac{n^3+n+4}{n^3+2n^2}= n^2 \log (1 - O(\frac{1}{n})) \rightarrow -\infty $$ Hence $$\lim_{n \rightarrow \infty} f(n) = e^{-\infty} = 0$$

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I suppose that $n\to \infty$

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

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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

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