Question

find f'(0) for f(x)= e^-1/x^2, x cannot equal 0
f(x)=x, x=0

Answer 1

looks like you have a piecewise function
for the derivative to exist at x=0 the function must be continuous and smooth first x=0

Answer 2

so we should check that first

Answer 3

\[\lim_{x \rightarrow 0}\frac{e^{-1}}{x^2}=0 ? \text{ does this equality hold } \]

Answer 4

is that the right function?

Answer 5

\[f(x)=\frac{e^{-1}}{x^2} ,x \neq 0 \\ f(x)=x, x=0\]

Answer 6

sorry, it's e^(-1/x^2)

Answer 7

\[\lim_{x \rightarrow a}f(x)=f(a) \text{ we are checking this right here for the continuity part } \\ \lim_{x \rightarrow 0}e^{\frac{-1}{x^2}}=0\]

Answer 8

does that equality hold?

Answer 9

like you can do a substitution on the left hand side
let u=-1/x^2
if x->0 then u->-infty
\[\lim_{u \rightarrow -\infty}e^{u}=?\]

Answer 10

I'm confused.. lol

Answer 11

So we have to check if the left side equal the right side at x=0?

Answer 12

we are checking the continuity part:
\[\lim_{x \rightarrow a}f(x)=f(a) \text{ <--we are checking this } \\ \lim_{x \rightarrow 0}f(x)=f(0) \text{ we want to see if this equality holds } \]

Answer 13

\[\lim_{x \rightarrow 0}e^{\frac{-1}{x^2}}=0 \text{ so we are checking to see if this equality holds }\]

Answer 14

we need to verify the left hand side does in fact equal the right hand side