Solving $\lim_{x\to 0}(\frac1x)(\frac1{\sqrt{1+x^{-2}}})$

Question

I'm trying to solve $\lim_{x\to 0-}(\frac1x)(\frac1{\sqrt{1+x^{-2}}})$ and $\lim_{x\to 0+}(\frac1x)(\frac1{\sqrt{1+x^{-2}}})$.

I've graphed the function so I do know that their respective values are -1 and 1, but I can't find these answers by simplifying and calculating no matter how hard I try. I guess the function could be transformed into $\frac{{{(1+\frac1{x^{2}}}})^{1/2}}{x}$, but I'm not sure if it's useful.

Can anyone help? Keep in mind that I'm not allowed to use L'Hopital.

Thank you.

For the second one the “limand“ is equal to $$\frac1{\sqrt{1+x^2}}$$ for the first one substitute $y=-x$. — Maximilian Janisch, Oct 15 '21 at 15:58
I don't really understand how to arrive to these two conclusions and why they're valid. — energyup, Oct 15 '21 at 16:01
I am using that if $x,y\ge 0$ then $$x \sqrt y = \sqrt{x^2y}.$$ — Maximilian Janisch, Oct 15 '21 at 16:03

score 2 · Accepted Answer · answered Oct 15 '21 at 16:09

2

$$\lim_{x \to 0} \dfrac{1}{x \sqrt {1+\dfrac{1}{x^2}}}=\lim_{x \to 0} \dfrac{\bbox[5px,border:2px solid red]{|x|}}{x \sqrt {x^2+1}}$$ $$\lim_{x \to 0^+}=\dfrac{x}{x \sqrt {x^2+1}}=1$$ $$\lim_{x \to 0^-}=\dfrac{-x}{x \sqrt {x^2+1}}=-1$$

answered Oct 15 '21 at 16:09

UNAN

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https://math.stackexchange.com/questions/486681/is-sqrtx2-always-pm-x – UNAN Oct 15 '21 at 16:51

Solving $\lim_{x\to 0}(\frac1x)(\frac1{\sqrt{1+x^{-2}}})$

1 Answers1