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log(1+x^2)
Limit of the function/ log(1+x^2)

Limit of the function log(1+x^2)

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        /     2\
 lim log\1 + x /
x->oo
lim⁡x→∞log⁡(x2+1)\lim_{x \to \infty} \log{\left(x^{2} + 1 \right)} 
Limit(log(1 + x^2), x, oo, dir='-')
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
02468-8-6-4-2-101005
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Other limits x→0, -oo, +oo, 1
lim⁡x→∞log⁡(x2+1)=∞\lim_{x \to \infty} \log{\left(x^{2} + 1 \right)} = \infty
lim⁡x→0−log⁡(x2+1)=0\lim_{x \to 0^-} \log{\left(x^{2} + 1 \right)} = 0
More at x→0 from the left
lim⁡x→0+log⁡(x2+1)=0\lim_{x \to 0^+} \log{\left(x^{2} + 1 \right)} = 0
More at x→0 from the right
lim⁡x→1−log⁡(x2+1)=log⁡(2)\lim_{x \to 1^-} \log{\left(x^{2} + 1 \right)} = \log{\left(2 \right)}
More at x→1 from the left
lim⁡x→1+log⁡(x2+1)=log⁡(2)\lim_{x \to 1^+} \log{\left(x^{2} + 1 \right)} = \log{\left(2 \right)}
More at x→1 from the right
lim⁡x→−∞log⁡(x2+1)=∞\lim_{x \to -\infty} \log{\left(x^{2} + 1 \right)} = \infty
More at x→-oo
Rapid solution [src] 
oo
∞\infty
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Limit of the function log(1+x^2)
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