lim x-0 (sin8x/x)

Question

lim x->0 (sin8x/x)

tkhunny · Answer

Think on this $\dfrac{\sin(8x)}{x} = 8\dfrac{\sin(8x)}{8x}$

anonymous · Answer

ok thanks, I've gotten it down to lim x->0 (8sin) so I think the answer is 8 but I don't know why

anonymous · Answer

I thought that the answer would be 0 since lim x->0 of sin is 0 and 0 multiplied by 8 would yield 0

tkhunny · Answer

"8sin"?  Oh, please say you did not do that!  That will get you kicked out of calculus so you can go back to Algebra 1!!  Seriously, that was very bad.

You shoudl have this $\lim\limits_{x\rightarrow 0}\dfrac{sin(x)}{x} = 1$.  It is a fundamental idea - almost treated as axiomatic in some cases.

anonymous · Answer

i just started taking it this semester. I'll try to avoid that, thanks for the advice! So would it be safe to treat the problem as lim x->0 sinx/x * lim x->0 8?

tkhunny · Answer

No, you're not quite seeing it.  More generally, as long as thing exist, we have this relatively general result.

$\lim\limits_{x\rightarrow 0}\dfrac{\sin(Qx)}{Qx} = 1$ for any constant, Q

$\lim\limits_{x\rightarrow 0}\dfrac{\sin(x)}{x} = 1$

$\lim\limits_{x\rightarrow 0}\dfrac{\sin(3x)}{3x} = 1$

$\lim\limits_{x\rightarrow 0}\dfrac{\sin(\sqrt{2}x)}{\sqrt{2}x} = 1$

$\lim\limits_{x\rightarrow 0}\dfrac{\sin(Pittsburgh(x))}{Pittsburgh(x)} = 1$

tkhunny · Answer

Now, go take another look at the first post I wrote.  You should see it.

anonymous · Answer

HAHAHAHAHA

anonymous · Answer

its so obvious now!! Thanks a million

tkhunny · Answer

Pittsburgh did it?

anonymous · Answer

yup...lol
all Pittsburgh