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log((1+x)/(1-x))
Limit of the function/ log((1+x)/(1-x))

Limit of the function log((1+x)/(1-x))

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You have entered [src] 
        /1 + x\
 lim log|-----|
x->oo   \1 - x/
lim⁡x→∞log⁡(x+11−x)\lim_{x \to \infty} \log{\left(\frac{x + 1}{1 - x} \right)} 
Limit(log((1 + x)/(1 - x)), 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
02468-8-6-4-2-10105-5
Plot the graph
Other limits x→0, -oo, +oo, 1
lim⁡x→∞log⁡(x+11−x)=iπ\lim_{x \to \infty} \log{\left(\frac{x + 1}{1 - x} \right)} = i \pi
lim⁡x→0−log⁡(x+11−x)=0\lim_{x \to 0^-} \log{\left(\frac{x + 1}{1 - x} \right)} = 0
More at x→0 from the left
lim⁡x→0+log⁡(x+11−x)=0\lim_{x \to 0^+} \log{\left(\frac{x + 1}{1 - x} \right)} = 0
More at x→0 from the right
lim⁡x→1−log⁡(x+11−x)=∞\lim_{x \to 1^-} \log{\left(\frac{x + 1}{1 - x} \right)} = \infty
More at x→1 from the left
lim⁡x→1+log⁡(x+11−x)=∞\lim_{x \to 1^+} \log{\left(\frac{x + 1}{1 - x} \right)} = \infty
More at x→1 from the right
lim⁡x→−∞log⁡(x+11−x)=iπ\lim_{x \to -\infty} \log{\left(\frac{x + 1}{1 - x} \right)} = i \pi
More at x→-oo
Rapid solution [src] 
pi*I
iπi \pi
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The graph
Limit of the function log((1+x)/(1-x))
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