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Why is $$\lim_{n\rightarrow\infty} n(1+{1\over n})^p$$ the same as $$\lim_{n\rightarrow\infty} n(1+{p\over n})$$ ?

I saw this answer in reference to a limit question here, and I am not sure how exactly to prove this to myself. Does it have something to do with the binomial theorem? This is in reference to $$\lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right).$$ $$\lim_{x\to-\infty}\sqrt{x^2+2x}\color{red}{+}\sqrt[3]{x^3+x^2}\\=\lim_{x\to-\infty}|x|\left(1+{2\over x}\right)^{1\over 2}\color{red}{+}x\left(1+{1\over x}\right)^{1\over3}\\=\lim_{x\to-\infty}|x|\left(1+{1\over 2}\cdot{2\over x}\right)\color{red}{+}x\left(1+{1\over 3}\cdot{1\over x}\right)=-{2\over 3}$$ I understand that the two both equal to infinity but it seems strange to just randomly multiply by the exponent and forget about it. Furthermore if it can be any number of my choice then would the limit still hold?

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    $\begingroup$ Both limits are $\infty$; there's no particular reason, not more than for $\lim_{n\to\infty}n=\infty=\lim_{n\to\infty}n^2$ $\endgroup$ – egreg Dec 12 '16 at 0:01
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What was used in the limits that motivated the question is that $$ (1+x)^a=1+ax+O(x^2) $$ so that $$ ((1+x)^a-(1+x)^b)=x·(a-b+O(x)) $$ Notice that the constant terms $1$ cancel each other. Thus the proper limit in the question should be $$ \lim_{n\to\infty}n·\left(\left(1+\frac1n\right)^p-1\right)=p. $$

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  • $\begingroup$ Thank you!!!! This was the question; the other answerers didn't seem to understand. Does what you have written have a name so I could read more about it? $\endgroup$ – cgug123 Dec 12 '16 at 14:40
  • $\begingroup$ This is just the linear Taylor polynomial with remainder term. The full expansion is the binomial series $(1+x)^a=1+ax+\frac12a(a-1)x^2+\frac16a(a-1)(a-2)x^3+…$. You could also consider $f(x)=(1+x)^a$, then $\lim_{x\to 0}\frac1x((1+x)^a-1)=f'(0)=a$ is the usual differential quotient. $\endgroup$ – LutzL Dec 12 '16 at 15:08
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No need for the binomial theorem. Since $x \mapsto x^p$ is continuous over $[0,\infty)$, then $$ \lim_{n \to \infty} \left(1+\frac1n \right)^p =1 $$ giving

$$ \lim_{n \to \infty} n\left(1+\frac1n \right)^p =\infty \cdot 1=\infty. $$

One also has

$$ \lim_{n \to \infty} n\left(1+{p\over n}\right)=\infty \cdot 1=\infty. $$

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  • $\begingroup$ Okay I understand that reasoning but then must it be p that becomes the numerator over n? Or could you just put in any number? $\endgroup$ – cgug123 Dec 12 '16 at 0:14
  • $\begingroup$ @cgug123 $(1+\frac{x}{\infty})^p = 1$ in the context of limits. $\endgroup$ – Natecat Dec 12 '16 at 0:18
  • $\begingroup$ @cgug123 You could put any number in the numerator, as long as it is a fixed number and doesn't depend on $n$. The limit of the expression in the bracket is still $1$, and the limit for the total expression is still $\infty$. $\endgroup$ – Arthur Dec 12 '16 at 0:21
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It's easy to see that the limits are the same.

I could add that since $$\lim_{n\rightarrow \infty} \frac{n(1+\frac{1}{n})^p}{n(1+\frac{p}{n})} =\lim_{n\rightarrow \infty} \frac{ (1+\frac{1}{n})^p}{ (1+\frac{p}{n})} = 1$$ we conclude that $n(1+\frac{1}{n})^p$ and $n(1+\frac{p}{n})$ are asymptotically equivalent, which is a bit stronger than simply having the same limit.

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