Recent content by dincerekin

there should be a negative sign before this integral and the charge is negative. two negatives make a positive \text{$\Delta $V}=\int _{\infty }^a\frac{1}{4\pi \epsilon }\frac{Q}{r^{2}}dr

well, i thought my answer was too simple/easy to be right i guess.

Homework Statement A thin plastic spherical shell of radius a is rubbed all over with wool and gains a charge of -Q. What is the potential relative to infinity at location B, a distance a/3 from the centre of the sphere?Homework Equations \text{$\Delta $V}=\int...

oh! how didn't I see this before thanks so much <3

it should be \lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}}=\lim_{x\to0}\frac{x^{-1}x^{2-k}}{e^\frac{1}{x^2}}

oh sorry, that should be the other way around

so, \lim_{x\to0}\frac{x^{2-k}}{e^\frac{1}{x^2}}=\lim_{x\to0}\frac{x^{-1}x^{1-k}}{e^\frac{1}{x^2}} but i can't simply say that \lim_{x\to0}\frac{x^{-1}x^{1-k}}{e^\frac{1}{x^2}}= \lim_{x\to0}{x^{-1}} × \lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}} because \lim_{x\to0}{x^{-1}} doesn't...

so I am trying to show its true for n=k+1 assuming n=k i.e i need to show that \lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}}=0 assuming that \lim_{x\to0}\frac{x^{2-k}}{e^\frac{1}{x^2}}=0 im not sure how to manipulate this now?

I applied l'hospital's and simplified a bit and now I've got \frac{n}{2}\lim_{x\to0}\frac{x^2}{e^\frac{1}{x^2}x^n}=0 now what?

Homework Statement Prove that \lim_{x\to0}\frac{e^\frac{-1}{x^2}}{x^n}=0 for any positive integer n. Homework Equations The Attempt at a Solution I've tried using a combination of induction and l'hopital's rule to no avail. Perhaps I am over complicating it? All help is...