Understanding Limits: Does it Exist or Go to Infinity?

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How can you tell if a limit does not exit or that it goes to infinity?

examplelim\underbrace{x\rightarrow}_{0}(\frac{\sqrt{x+1}}{x})

The x goes to 0The book says the limit is \infty but if you take the left side limit you get -\infty and if you take the right side of the limit you \infty. So wouldn't this limit not exist, because you have two different limits?
 
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Yes, I agree with you. I think there's a typo somewhere.
 
That's what I thought, thank you for looking at it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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