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?

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?

anonymous · Answer

substitute mx into the first expression (for h)

anonymous · Answer

and then? I don't understand why

turingtest · Answer

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

anonymous · Answer

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

turingtest · Answer

sorry...

turingtest · Answer

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

turingtest · Answer

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

anonymous · Answer

I got it! Thanks

turingtest · Answer

welcome