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I have the following function: $$ f(x)=e^{x^2}-1-x^2 $$ It is considered known that $ f(x) \geq \frac{x^4}{2} $ for any real $x$'s. I have to check the differentiability of the function $ g(x)=f(x)^{1/4} $ at the point $ x=0 $.

The derivative of $ g $ is $\frac{x(e^{x^2})}{2(e^{x^2}-x^2-1)^{3/4}} $. I tried computing the lateral limits, using either the known fact I mentioned above, L'Hospital, or the fact that $e^x>x+1$, but I can't seem to manage to do it.

The inequality that I mentioned was proved earlier in the problem, and I'm not sure if it is supposed to be used here as well. Thanks for your help!

Wolfuryo
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  • whenever you have a function in the form $ \dfrac{e^{f(x)}}{g(x)} $, then exponential function increases or decreases rapidly than that of polynomial function. Use this to find limits. – Manjoy Das Apr 29 '20 at 20:11
  • Okay, I'll keep that in mind! – Wolfuryo Apr 29 '20 at 20:11

1 Answers1

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See that for $f(x)=e^{x^2}-1-x^2$, $g(x)=(f(x))^{\frac{1}{4}}$ ; $g(0)=0$

So $g'(x)=\lim_{x\to0}{\frac{g(x)-g(0)}{x}}$, provided the limit exists. $\; \Rightarrow g'(x)=\lim_{x\to0}{\frac{g(x)}{x}}$

$\; \Rightarrow g'(x)=\lim_{x\to0}{\frac{(e^{x^2}-1-x^2)^\frac{1}{4}}{x}}$

Use Taylor's series expansion of $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ to get $e^{x^2}-1-x^2=\frac{x^4}{2!}+\frac{x^6}{3!}+...$

and $g'(x)=\lim_{x\to0}{(\frac{\frac{x^4}{2!}+\frac{x^6}{3!}+...}{x^4})^{1/4}}=\lim_{x\to0}{(\frac{1}{2!}+\frac{x^2}{3!}+...)^{1/4}}=(\frac{1}{2})^{1/4}.$

Thus $g'(x)$ exists and equals $(\frac{1}{2})^{1/4}$. So $g(x)$ is differentiable at $0$.

amWhy
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udit narayan pandey
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  • Thanks! I'll keep the question open since this is a bit beyond what we were supposed to know. – Wolfuryo Apr 29 '20 at 20:09
  • When you put the denominator(the x) under the power, if you take the left limit(x negative, going towards 0), you would have had to add a -. This would lead to different lateral limits, therefore the function would not differentiable. – Wolfuryo Apr 30 '20 at 13:02
  • I think you are right. – udit narayan pandey May 01 '20 at 02:26
  • Since @Wolfuryo poitned this out, it would be better if you edit the answer.. – udit narayan pandey May 01 '20 at 02:33
  • I made the edit. I'm not sure I explained everything properly, so feel free to make modifications if you want. I'll mark this answer as accepted once the edit is approved. – Wolfuryo May 01 '20 at 12:17