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

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

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 lim  log\x + \/  1 + x  /
x->-oo
lim⁡x→−∞log⁡(x+x2+1)\lim_{x \to -\infty} \log{\left(x + \sqrt{x^{2} + 1} \right)} 
Limit(log(x + sqrt(1 + x^2)), x, -oo)
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+x2+1)=−∞\lim_{x \to -\infty} \log{\left(x + \sqrt{x^{2} + 1} \right)} = -\infty
lim⁡x→∞log⁡(x+x2+1)=∞\lim_{x \to \infty} \log{\left(x + \sqrt{x^{2} + 1} \right)} = \infty
More at x→oo
lim⁡x→0−log⁡(x+x2+1)=0\lim_{x \to 0^-} \log{\left(x + \sqrt{x^{2} + 1} \right)} = 0
More at x→0 from the left
lim⁡x→0+log⁡(x+x2+1)=0\lim_{x \to 0^+} \log{\left(x + \sqrt{x^{2} + 1} \right)} = 0
More at x→0 from the right
lim⁡x→1−log⁡(x+x2+1)=log⁡(1+2)\lim_{x \to 1^-} \log{\left(x + \sqrt{x^{2} + 1} \right)} = \log{\left(1 + \sqrt{2} \right)}
More at x→1 from the left
lim⁡x→1+log⁡(x+x2+1)=log⁡(1+2)\lim_{x \to 1^+} \log{\left(x + \sqrt{x^{2} + 1} \right)} = \log{\left(1 + \sqrt{2} \right)}
More at x→1 from the right
Rapid solution [src] 
-oo
−∞-\infty
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The graph
Limit of the function log(x+sqrt(1+x^2))
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