Evaluate the following limits, assuming all angles are in radian

a) lim x->0 (sin5x)/(sin6x)

b) lim x->0 (xsin6x)/(sin^2(9x))

c) lim x->0 (sin3x)/(9x-2tanx)

2. do you know l'hospital's rule?

3. yeah i know l'hopital's rule...i definitely did not understand the second part you said but if i use l'hopital it should work

4. Originally Posted by katekate
yeah i know l'hopital's rule...i definitely did not understand the second part you said but if i use l'hopital it should work

lol that's just my signature and has nothing to do with you or this post, but let me do the first one and have you try the second one

$\lim_{x\rightarrow 0}\frac{\sin(5x)}{\sin(6x)}$ is indeterminant, meaning you get $\frac{0}{0}$ if you try to plug into the expression, so we apply L'hospital's rule to get

$\lim_{x\rightarrow 0}\frac{\sin(5x)}{\sin(6x)}=\lim_{x\rightarrow 0}\frac{5\cos(5x)}{6\cos(6x)}=\frac{5\cos(0)}{6\co s(0)}=\frac{5}{6}$

do the same thing for the next

5. I can't get the second one but I get 3/5 for the third. I think im messing up the derivatives of both top and bottom.

Edit: never mind 3/5 is definitely wrong too

6. Originally Posted by katekate
I can't get the second one but I get 3/5 for the third. I think im messing up the derivatives of both top and bottom.

Edit: never mind 3/5 is definitely wrong too

alright so let's do this

LH means i used l'hospital's rule

$\lim_{x\rightarrow 0}\frac{x\sin(6x)}{\sin^2(9x)}=LH=\lim_{x\rightarr ow 0}\frac{\sin(6x)+6x\cos(6x)}{2\cdot 9\sin(9x)\cos(9x)}$

Since we still get $\frac{0}{0}$ we use LH again

$=\lim_{x\rightarrow 0}\frac{6\cos(6x)+6\cos(6x)-36x\sin(6x)}{162\cos^2(9x)-162\sin^2(9x)}=\frac{12\cos(0)}{162\cos^2(0)}=\fra c{12}{162}=\frac{2}{27}$

any questions?

try the third one more time before I help you out there too

7. great i got both of them now, i was squaring the 2 in the denominator when i shouldn't have been before in the third one. i didnt realize i had to use l'hopitals more than once, thanks a ton for everything