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n^2+n^3

Limit of the function n^2+n^3

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     / 2    3\
 lim \n  + n /
n->oo         
$$\lim_{n \to \infty}\left(n^{3} + n^{2}\right)$$
Limit(n^2 + n^3, n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(n^{3} + n^{2}\right)$$
Let's divide numerator and denominator by n^3:
$$\lim_{n \to \infty}\left(n^{3} + n^{2}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\frac{1}{n^{3}}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{1 + \frac{1}{n}}{\frac{1}{n^{3}}}\right) = \lim_{u \to 0^+}\left(\frac{u + 1}{u^{3}}\right)$$
=
$$\frac{1}{0} = \infty$$

The final answer:
$$\lim_{n \to \infty}\left(n^{3} + n^{2}\right) = \infty$$
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
Rapid solution [src]
oo
$$\infty$$
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty}\left(n^{3} + n^{2}\right) = \infty$$
$$\lim_{n \to 0^-}\left(n^{3} + n^{2}\right) = 0$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(n^{3} + n^{2}\right) = 0$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(n^{3} + n^{2}\right) = 2$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(n^{3} + n^{2}\right) = 2$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(n^{3} + n^{2}\right) = -\infty$$
More at n→-oo
The graph
Limit of the function n^2+n^3