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Evaluate the Limit $$L=\lim_{x \to 0}\frac{x}{\sqrt{1-e^{-x^2}}}$$

Now it is in Indeterminate form $\frac{0}{0}$

I Tried using L'Hopital's Rule as below:

$$L=\lim_{x \to 0}\frac{1}{\frac{1}{2\sqrt{1-e^{-x^2}}}{\left(-e^{-x^2}\right)}{(-2x)}}$$

$\implies$

$$L=\lim_{x \to 0}\frac{\sqrt{1-e^{-x^2}}}{x}=\frac{1}{L}$$

hence

$$L=1$$

is this right approach?

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    $\begingroup$ You showed that if the limit at $0^+$ exists then it is equal to $1$. $\endgroup$ – Robert Z Jun 14 '18 at 18:30
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    $\begingroup$ It's also not correct as $L = 1/L \implies L^2 = 1$ so not $L=1$ but $L=1$ or $L=-1$ (in fact both of these are limits of subsecuences) $\endgroup$ – Winther Jun 14 '18 at 18:30
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The function is tricky since

$$ \lim_{x\to \color{blue}{0^+}} \frac{x}{\sqrt{1 - e^{-x^2}}} = \color{blue}{+1} $$

and

$$ \lim_{x\to \color{red}{0^-}} \frac{x}{\sqrt{1 - e^{-x^2}}} = \color{red}{-1} $$

So, technically

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

does not exist

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  • $\begingroup$ Indeed. Shown here $\endgroup$ – Rhys Hughes Jun 14 '18 at 18:40
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You have shown quite correctly that the limit, if it exists, must satisfy the equation $L=1/L$. On the other hand, the function $x/\sqrt{1-e^{-x^2}}$ is odd, and therefore, the (two-sided) limit, if it exists, can only be $0$. Since $L=0$ does not satisfy the equation $L=1/L$, we can conclude that the limit does not exist!

The take-home message here is to keep in mind a crucial part of the statement of L'Hopital's Rule: It says that $\lim_{x\to c}(f(x)/g(x))=\lim_{x\to c}(f'(x)/g'(x))$ if the latter limit exists. If $\lim_{x\to c}(f'(x)/g'(x))$ does not exist, L'Hopital's Rule is silent.

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If we consider your limit as a one sided limit for now (say we approach $0$ from the right), we can write the function as $\sqrt{\frac{x^2}{1-e^{-x^2}}}$ and applying L'Hospital inside the radical goes very easy, since the $2x$ in the numerator will be canceled against an emerging $2x$ from the denominator (Chainrule). Since $\sqrt{x^2}=|x|$, the left sided limit can be addressed as well.

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L'Hopital is a little tricky and inconclusive here. For one thing, as commented, you only "concluded" that $L^2=1$.

Are you allowed to use series? If so, around $x=0$ we have

$$ e^{-x^2} = 1 - x^2 + o(x^2)$$

then

$$ \lim_{x \to 0}\frac{x}{\sqrt{1-e^{-x^2}}} = \lim_{x \to 0} \frac{x}{\sqrt{x^2}}= \lim_{x \to 0} \frac{x}{|x|}= \lim_{x \to 0} {\rm sign}(x) $$

But the lateral limits of the sign function are $1$ and $-1$, hence the limit does not exist.

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