Limits of derivatives of an exponential

[Catria]

Homework Statement

Determine the lowest derivative order for which the limit towards 0+ of the nth order derivative of f is nonzero (or otherwise does not exist). f = e^{\frac{-1}{x^{2}}}

Homework Equations

lim_{x\rightarrow0+}\frac{d^{n}}{dx^{n}}e^{\frac{-1}{x^{2}}}

The Attempt at a Solution

lim_{x\rightarrow0+}\frac{d}{dx}e^{\frac{-1}{x^{2}}} = 0

lim_{x\rightarrow0+}\frac{d^{2}}{dx^{2}}e^{\frac{-1}{x^{2}}} = 0

lim_{x\rightarrow0+}\frac{d^{3}}{dx^{3}}e^{\frac{-1}{x^{2}}} = 0

 

[SammyS]

Homework Statement

Determine the lowest derivative order for which the limit towards 0+ of the nth order derivative of f is nonzero (or otherwise does not exist). f = e^{\frac{-1}{x^{2}}}

Homework Equations

lim_{x\rightarrow0+}\frac{d^{n}}{dx^{n}}e^{\frac{-1}{x^{2}}}

The Attempt at a Solution

lim_{x\rightarrow0+}\frac{d}{dx}e^{\frac{-1}{x^{2}}} = 0

lim_{x\rightarrow0+}\frac{d^{2}}{dx^{2}}e^{\frac{-1}{x^{2}}} = 0

lim_{x\rightarrow0+}\frac{d^{3}}{dx^{3}}e^{\frac{-1}{x^{2}}} = 0

Use the chain rule.

\displaystyle \frac{d}{dx}e^{-1/x^2}= \frac{2e^{-1/x^2}}{x^3}\ .

 

[Catria]

You're not differentiating correctly.

Use the chain rule !

\displaystyle \frac{d}{dx}e^{-1/x^2}= \frac{2e^{-1/x^2}}{x^3}\ .

I tried that at the first three orders and I still had the limit of these derivatives towards 0+ as 0.

 

[SammyS]

I tried that at the first three orders and I still had the limit of these derivatives towards 0+ as 0.

Right!

I get zero for the fourth derivative also.

I don't see how it will ever be anything else, no matter how high the order of the derivative, but I haven't proved that to myself.

.

 

[HallsofIvy]

Any derivative is e^{-1/x^2} over a polynomial and its limit as x goes to 0 will always be 0.