How to show f′(0) exists for $f\left(x\right)={e}^{\frac{-1}{{x}^{2}}}\mathrm{sin}\left(\frac{1}{x}\right)$ for $x\ne 0$ and f(0)=0?

Michael Maggard
2022-01-14
Answered

How to show f′(0) exists for $f\left(x\right)={e}^{\frac{-1}{{x}^{2}}}\mathrm{sin}\left(\frac{1}{x}\right)$ for $x\ne 0$ and f(0)=0?

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Hattie Schaeffer

Answered 2022-01-15
Author has **37** answers

Hint

$\left|\frac{{e}^{\frac{-1}{{h}^{2}}}\mathrm{sin}\left(\frac{1}{h}\right)}{h}\right|\le \left|\frac{{e}^{-\frac{1}{{h}^{2}}}}{h}\right|$

$\underset{h\to 0}{lim}\frac{{e}^{-\frac{1}{{h}^{2}}}}{\left|h\right|}$

can be calculated with the substritution$x=\frac{1}{\left|h\right|}$

can be calculated with the substritution

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My Attempt

$\mathrm{\Delta}={\mathrm{cos}}^{2}p-4(\mathrm{cos}p-1)\mathrm{sin}p\ge 0$

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I first started off on the LHS and managed to get the denominator to become$1-{\mathrm{cot}}^{2}x$ by multiplying by $\mathrm{sin}x-\mathrm{cos}x$ and then dividing by ${\mathrm{sin}}^{2}x$ , but from there I had no idea how to continue.

I first started off on the LHS and managed to get the denominator to become