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Limit of the function
:
Limit of 1-cos(3*x)
Limit of ((6+5*x)/(-1+5*x))^((1+2*x^2)/x)
Limit of (-3*5^(1+x)+5*2^x)/(2*5^x+100*2^x)
Limit of (4^x-9^x)/(4^(1+x)+9^(1+x))
Sum of series
:
n^2+n^3
Identical expressions
n^ two +n^ three
n squared plus n cubed
n to the power of two plus n to the power of three
n2+n3
n²+n³
n to the power of 2+n to the power of 3
Similar expressions
n^2-n^3
Limit of the function
/
n^2+n^3
Limit of the function n^2+n^3
at
→
Calculate the limit!
v
For end points:
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From the left (x0-)
From the right (x0+)
The graph:
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Piecewise:
{
enter the piecewise function here
The solution
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[src]
/ 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
Plot the graph
Rapid solution
[src]
oo
$$\infty$$
Expand and simplify
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