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

I have this limit and I don't have the slightest idea about how to solve it. I know that it doesn't exist because the right-hand limit is $1$ and the left-hand limit is $-1$. However, I'm not sure about how to solve it without actually computing values close to zero. I already tried factoring to eliminate the discontinuity, I also tried L'Hôpital's rule, tried algebraic manipulation (i.e. squaring the function) and I don't know what else to try.

Additionally, I would be very thankful if you could point me to resources, or book chapters where I can learn how to deal with harder limits (not necessarily including trigonometric functions, as I study economics and we barely deal with that kind of functions at all.)

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$$\\ \lim _{ x\rightarrow 0 } x\sqrt { 1+\frac { 1 }{ x^{ 2 } } } =\lim _{ x\rightarrow 0 } x\frac { \sqrt { 1+x^{ 2 } } }{ \left| x \right| } =\lim _{ x\rightarrow 0 } \begin{cases} \sqrt { 1+{ x }^{ 2 } } & x>0 \\ -\sqrt { 1+{ x }^{ 2 } } &x<0 \end{cases}=\lim _{ x\rightarrow 0 } \begin{cases} 1& x>0 \\ -1&x<0 \end{cases}\\ $$

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  • $\begingroup$ I took the liberty of adding & to your cases environment for improved readability. I hope you don't mind. $\endgroup$ – Reveillark Sep 9 '16 at 0:44
  • $\begingroup$ What is the logic for dividing the whole function by $\left |x \right|$ in your second step without changing the resulting limit? I don't understand how you go from step 1 to step 2, but if I take that step for granted, the rest is easy to follow. $\endgroup$ – Maximiliano Santiago Sep 9 '16 at 1:00
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    $\begingroup$ @MaximilianoSantiago,$\\ x\sqrt { 1+\frac { 1 }{ x^{ 2 } } } =x\sqrt { \frac { 1+{ x }^{ 2 } }{ { x }^{ 2 } } } =x\frac { \sqrt { 1+{ x }^{ 2 } } }{ \sqrt { { x }^{ 2 } } } =x\frac { \sqrt { 1+{ x }^{ 2 } } }{ \left| x \right| } \\ \\ \\ $ $\endgroup$ – haqnatural Sep 9 '16 at 1:04
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Note that you should move the $x$ into the radical:

$$\begin{array} \ x\sqrt{1+x^{-2}} & = \sqrt{x^2}\sqrt{1+x^{-2}} \\ & = \sqrt{x^2(1+x^{-2})} \\ & = \sqrt{x^2+1} \\ \end{array}$$

if $x>0$

If $x<0$,

$$x\sqrt{1+x^{-2}}=-\sqrt{x^2+1}$$

Then the limit should be very easy.

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