Attachment 25379

Does this require l'Hospitals's Rule or is there another way to do this?

l'hospital's rule would give lim x->0 cos(x^2)/1 which is = 1/1=1

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- October 23rd 2012, 09:17 PMkingsolomonsgravelimit as x goes to zero of sin(x^2)/x
Attachment 25379

Does this require l'Hospitals's Rule or is there another way to do this?

l'hospital's rule would give lim x->0 cos(x^2)/1 which is = 1/1=1 - October 23rd 2012, 09:42 PMMarkFLRe: limit as x goes to zero of sin(x^2)/x
You are forgetting to apply the chain rule in the numerator...you could also multiply by x/x to get the same result of 0.

- October 23rd 2012, 10:10 PMkingsolomonsgraveRe: limit as x goes to zero of sin(x^2)/x
so for the the above it should be which would be 1*0/1=0

multiplying by x/x we would get which is equal is also equal to zero too but how? I feel like i'm missing some obvious step for version where we multiply by x/x

also is ? or is it just the last two that are the same? - October 23rd 2012, 10:20 PMProve ItRe: limit as x goes to zero of sin(x^2)/x
- October 23rd 2012, 10:20 PMMarkFLRe: limit as x goes to zero of sin(x^2)/x
We have:

For the second limit let and as we have so we may write:

I am assuming you have been given to use for such limits. - October 23rd 2012, 10:25 PMkingsolomonsgraveRe: limit as x goes to zero of sin(x^2)/x
Thanks, yes we were given lim as x-> 0 sin(u)/u = 1

thanks much