Question

Use the result that the limit as h->0 of (1-cos h)/h^2 = 1/2 to prove that for m not equal to 0, lim as x-> 0 of (cos mx-1)/x^2= -m^2/2. What happens when m=0?

Answer 1

substitute mx into the first expression (for h)

Answer 2

and then? I don't understand why

Answer 3

\[\lim_{x\to0}\frac{1-\cos(mx)}{x^2}=-m^2\lim_{x\to0}\frac{\cos(mx)-1}{m^2x^2}\]

Answer 4

but the original equation is cosmx-1/x^2

Answer 5

sorry...

Answer 6

\[\lim_{x\to0}\frac{\cos(mx)-1}{x^2}=-m^2\lim_{x\to0}\frac{1-\cos(mx)}{m^2x^2}\]

Answer 7

okay, NOW you should be able to take it from here lol

Answer 8

I got it! Thanks

Answer 9

welcome