So I know how to take the general derivative of this equation. It's a simple product rule. I have that. My problem is, I need to show that the derivative at x=0 is 0. I know that I'm supposed to use this equation. f'(x)= lim x->0 of [f(x+h)-f(x)]/h So I plug in x+h and I get: [(e^(-1/(x+h)^2))-(e^(-1/x^2))]/h My problem is, I have no idea how to simplify that. I know that I need to get rid of the denominator, but I'm not sure how I can do that. Suggestions?
Thanks. That worked perfectly. I actually have another question now. I need to find a Taylor series for x^(1/2) about a general center c=a^2. I've been taking derivatives and trying to find a pattern. I seem to be on the verge of it but I just can quite get it. Anyone know what the formula for the kth derivative of f(x) might be?
This is a great solution, except for one logical error. In your case of the e^{a} expansion, a = 1/x^{2} as x -> 0. Therefore, a is in fact approaching infinity. So you cannot assume that the e^{a} expansion is equivalent to 1 + a + a^{2}/2 since a is not small. Instead, just substitute in the entire e^{a} expansion, then take the limit, and you will see that the denominator will equal 0 + 0 + ∞ + ∞ + ... Therefore the limit is still 2/∞ = 0.
Oh dear, I completely missed that part :( so no need to expand the exponential at all. Because exponentials increase a lot faster than polynomials decrease. So by inspection e^(x^(-2))*x^3 -> infinity , when x->0 Thanks p.s. L'Hopital's rule will not work is this case as well , :( as it requires (infinity/infinity) or (zero/zero) limit.