Why are these two limits equal? 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?
 A: 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
$$
\lim_{n\to\infty}n·\left(\left(1+\frac1n\right)^p-1\right)=p.
$$
A: 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

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

A: 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.
