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I'd like to get a step by step answer for the following:

$$\lim_{x \to 0^+} x \sqrt{1 + \frac 1 {x^2}}$$

Alexis
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  • Can you share what you've tried, and explain what you're having trouble with? Do you mean this? $$\lim_{x \to 0^+} x \sqrt{1 + \frac{1}{x^2}}$$ –  Jun 19 '14 at 02:41
  • Yes exactly, I tried multiplying both the numerator and denominator by (1+1/x^2)^(1/2) but I then get stuck on trying to clear it up – Alexis Jun 19 '14 at 02:45

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Hint: Note that for positive $x$ we have $$\sqrt{1+\frac{1}{x^2}}=\frac{1}{x}\sqrt{x^2+1}.$$ You can probably take over from here.

Remark: Note that if $x$ is negative, then $\sqrt{x^2}=|x|=-x$. So the limit from the left is $-1$.

André Nicolas
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  • Thanks, this would explain it. I still do not get that equality, probably a property of square roots I can't remember, I will check them out ! – Alexis Jun 19 '14 at 02:52
  • We have $1+\frac{1}{x^2}=\frac{x^2+1}{x^2}$ (brought to common denominator). And if $a$ and $b$ are positive, then $\sqrt{\frac{a}{b}}=\frac{1}{\sqrt{b}}\cdot\sqrt{a}$. – André Nicolas Jun 19 '14 at 02:55