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1/(1+n^2)
Limit of the function/ 1/(1+n^2)

Limit of the function 1/(1+n^2)

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The solution

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       1   
 lim ------
n->oo     2
     1 + n
lim⁡n→∞1n2+1\lim_{n \to \infty} \frac{1}{n^{2} + 1} 
Limit(1/(1 + n^2), n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞1n2+1\lim_{n \to \infty} \frac{1}{n^{2} + 1}
Let's divide numerator and denominator by n^2:
lim⁡n→∞1n2+1\lim_{n \to \infty} \frac{1}{n^{2} + 1} =
lim⁡n→∞(1n2(1+1n2))\lim_{n \to \infty}\left(\frac{1}{n^{2} \left(1 + \frac{1}{n^{2}}\right)}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞(1n2(1+1n2))=lim⁡u→0+(u2u2+1)\lim_{n \to \infty}\left(\frac{1}{n^{2} \left(1 + \frac{1}{n^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u^{2}}{u^{2} + 1}\right)
=
0202+1=0\frac{0^{2}}{0^{2} + 1} = 0

The final answer:
lim⁡n→∞1n2+1=0\lim_{n \to \infty} \frac{1}{n^{2} + 1} = 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
02468-8-6-4-2-101002
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Rapid solution [src] 
0
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Expand and simplify
Other limits n→0, -oo, +oo, 1
lim⁡n→∞1n2+1=0\lim_{n \to \infty} \frac{1}{n^{2} + 1} = 0
lim⁡n→0−1n2+1=1\lim_{n \to 0^-} \frac{1}{n^{2} + 1} = 1
More at n→0 from the left
lim⁡n→0+1n2+1=1\lim_{n \to 0^+} \frac{1}{n^{2} + 1} = 1
More at n→0 from the right
lim⁡n→1−1n2+1=12\lim_{n \to 1^-} \frac{1}{n^{2} + 1} = \frac{1}{2}
More at n→1 from the left
lim⁡n→1+1n2+1=12\lim_{n \to 1^+} \frac{1}{n^{2} + 1} = \frac{1}{2}
More at n→1 from the right
lim⁡n→−∞1n2+1=0\lim_{n \to -\infty} \frac{1}{n^{2} + 1} = 0
More at n→-oo
The graph
Limit of the function 1/(1+n^2)
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