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

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

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You have entered [src] 
        /     2\
 lim log\1 + t /
t->0+
lim⁡t→0+log⁡(t2+1)\lim_{t \to 0^+} \log{\left(t^{2} + 1 \right)} 
Limit(log(1 + t^2), t, 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
02468-8-6-4-2-10105-5
Plot the graph
Rapid solution [src] 
0
00
Expand and simplify
One‐sided limits [src] 
        /     2\
 lim log\1 + t /
t->0+
lim⁡t→0+log⁡(t2+1)\lim_{t \to 0^+} \log{\left(t^{2} + 1 \right)} 
0
00 
= -2.87136948736244e-33
        /     2\
 lim log\1 + t /
t->0-
lim⁡t→0−log⁡(t2+1)\lim_{t \to 0^-} \log{\left(t^{2} + 1 \right)} 
0
00 
= -2.87136948736244e-33
= -2.87136948736244e-33
Other limits t→0, -oo, +oo, 1
lim⁡t→0−log⁡(t2+1)=0\lim_{t \to 0^-} \log{\left(t^{2} + 1 \right)} = 0
More at t→0 from the left
lim⁡t→0+log⁡(t2+1)=0\lim_{t \to 0^+} \log{\left(t^{2} + 1 \right)} = 0
lim⁡t→∞log⁡(t2+1)=∞\lim_{t \to \infty} \log{\left(t^{2} + 1 \right)} = \infty
More at t→oo
lim⁡t→1−log⁡(t2+1)=log⁡(2)\lim_{t \to 1^-} \log{\left(t^{2} + 1 \right)} = \log{\left(2 \right)}
More at t→1 from the left
lim⁡t→1+log⁡(t2+1)=log⁡(2)\lim_{t \to 1^+} \log{\left(t^{2} + 1 \right)} = \log{\left(2 \right)}
More at t→1 from the right
lim⁡t→−∞log⁡(t2+1)=∞\lim_{t \to -\infty} \log{\left(t^{2} + 1 \right)} = \infty
More at t→-oo
Numerical answer [src] 
-2.87136948736244e-33
-2.87136948736244e-33
The graph
Limit of the function log(1+t^2)
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