Question

use l'hopital's rule to evaluate the limit \[\lim_{h \rightarrow 0}\frac{ \sin (x+h)-\sin x }{ h }\] x is a real number

Answer 1

@genius12 not sure what I would have to do for this one because there are two variables

Answer 2

Angle sum identity for sine:\[\sin(x+h)=\sin x\cos h + \sin h\cos x\]
The limit becomes
\[\lim_{h\to0}\frac{\sin x\cos h + \sin h\cos x-\sin x}{h}\\
\lim_{h\to0}\frac{\sin h\cos x}{h}+\lim_{h\to0}\frac{\sin x(\cos h-1)}{h}\\
\cos x\lim_{h\to0}\frac{\sin h}{h}+\sin x\lim_{h\to0}\frac{\cos h-1}{h}\]

Answer 3

ah I see. and then you just find the derivative and substitute 0 for h?

Answer 4

What you're doing with the initial limit is finding the derivative of \(\sin x\); no other derivative-taking involved.

Answer 5

What I mean to say is, you don't \(\textbf{need}\) to use L\Hopital's rule here. The remaining two limits are known trig limits, but you can use L'Hopital's if you like, yes.

Answer 6

but if you left it here you would still have 0/0 wouldn't you?

Answer 7

Yeah, if you were to just plug in h = 0. Since you have to use L'Hopital's rule, go ahead and use it. It really simplifies things.

Answer 8

well now that's where I'm confused. I thought you said you don't have to find any further derivative. Or was that because I didn't have to use his rule, but now that I am I should find another derivative?

Answer 9

I'll try to clear up my confusing explanation:

Do you know the following limits?
\[\lim_{x\to0}\frac{\sin(ax)}{ax}=1,\;\text{where }a\not=0\]
and
\[\lim_{x\to0}\frac{\cos(ax)-1}{ax}=0,\;\text{where }a\not=0\]
If you do, you'll notice that these are the two limits you have left. The last line of calculation thus reduces to \(\cos x,\) which is indeed the derivative of \(\sin x\). The reason it's the derivative is by the limit definition of the derivative of \(\sin x\):
\[\frac{d}{dx}\sin x=\color{red}{\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}}\]
The red limit is the limit you started off with.

But, you don't have to use the knowledge of the trig limits I listed above. You can evaluate the limits without being familiar with them (meaning you can find \(\displaystyle\lim_{h\to0}\frac{\sin{h}}{h}\)) by simply applying L'Hopital's rule.

Answer 10

ok that makes sense I'll have to look into those trig limits that would've saved a lot of trouble and confusion