Using the fact that lim h->0 sin(h)/h = 1 and lim h->0 cos(h)-1/h =0 compute the following limits a. lim h->0 sin(x+h)-sin(x)/h b lim h->0 cos(x+h)-cos(x)/h
addition angle formula for the first one second one too
\[\sin(x+h)-\sin(x)=\sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)\] divide all by \(h\) and take the limit
you get that?
I think how about b
exactly the same, but this time the addition angle formula for cosine
i can walk you through one if you like the other is the same
satelline, you will do all of that work?
btw do you know what the answer is before you begin? @SolomonZelman what work?
k please
to which one ? Also, I am logging off in 2 minutes... SCHEDULE :P
sorry
lets do the first one
k
I am not a good teacher anyway ... bye
better than me
before we begin, is it clear that you are being asked for the derivative of sine, and that the answer is cosine? to the first one i mean
yes
k lets ignore the limit for the moment so i don't have to write it in, and just to the algebra first
\[\frac{\sin(x+h)-\sin(x)}{h}=\frac{\sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)}{h}\] is a start that by the "addition angle formula for sine
now break in to two parts \[\frac{\sin(x)\cos(h)-\sin(x)}{h}+\frac{\sin(h)\cos(x)}{h}\] factor the first one and get \[\sin(x)\frac{\cos(h)-1)}{h}+\frac{\sin(h)\cos(x)}{h}\]
now take the limit as \(h\to 0\)
since \[\lim_{h\to 0}\frac{\cos(h)-1}{h}=0\] the first term goes bye bye and since \[\lim_{h\to 0}\frac{\sin(h)}{h}=1\] the second term becomes \(\cos(x)\) and needed
I think I got it but what formula for b?
the addition angle for cosine is \[\cos(x+h)=\cos(x)\cos(h)-\sin(x)\sin(h)\]
do you have time to show the work?
i have the time, but why don't you try it first if you mimic what i did above, you will a) understand it better and b) learn what it is about it you don't understand then ask
k thanks so much
yw write if you get stuck
Join our real-time social learning platform and learn together with your friends!