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2/n^2

Limit of the function 2/n^2

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The solution

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     /2 \
 lim |--|
n->oo| 2|
     \n /
limn(2n2)\lim_{n \to \infty}\left(\frac{2}{n^{2}}\right)
Limit(2/n^2, n, oo, dir='-')
Detail solution
Let's take the limit
limn(2n2)\lim_{n \to \infty}\left(\frac{2}{n^{2}}\right)
Let's divide numerator and denominator by n^2:
limn(2n2)\lim_{n \to \infty}\left(\frac{2}{n^{2}}\right) =
limn(21n21)\lim_{n \to \infty}\left(\frac{2 \frac{1}{n^{2}}}{1}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
limn(21n21)=limu0+(2u2)\lim_{n \to \infty}\left(\frac{2 \frac{1}{n^{2}}}{1}\right) = \lim_{u \to 0^+}\left(2 u^{2}\right)
=
202=02 \cdot 0^{2} = 0

The final answer:
limn(2n2)=0\lim_{n \to \infty}\left(\frac{2}{n^{2}}\right) = 0
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
02468-8-6-4-2-10100200
Rapid solution [src]
0
00
Other limits n→0, -oo, +oo, 1
limn(2n2)=0\lim_{n \to \infty}\left(\frac{2}{n^{2}}\right) = 0
limn0(2n2)=\lim_{n \to 0^-}\left(\frac{2}{n^{2}}\right) = \infty
More at n→0 from the left
limn0+(2n2)=\lim_{n \to 0^+}\left(\frac{2}{n^{2}}\right) = \infty
More at n→0 from the right
limn1(2n2)=2\lim_{n \to 1^-}\left(\frac{2}{n^{2}}\right) = 2
More at n→1 from the left
limn1+(2n2)=2\lim_{n \to 1^+}\left(\frac{2}{n^{2}}\right) = 2
More at n→1 from the right
limn(2n2)=0\lim_{n \to -\infty}\left(\frac{2}{n^{2}}\right) = 0
More at n→-oo
The graph
Limit of the function 2/n^2