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)
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
$\displaystyle \lim_{x\rightarrow 0}\frac{\sin(5x)}{\sin(6x)}$ is indeterminant, meaning you get $\displaystyle \frac{0}{0}$ if you try to plug into the expression, so we apply L'hospital's rule to get
$\displaystyle \lim_{x\rightarrow 0}\frac{\sin(5x)}{\sin(6x)}=\lim_{x\rightarrow 0}\frac{5\cos(5x)}{6\cos(6x)}=\frac{5\cos(0)}{6\co s(0)}=\frac{5}{6}$
do the same thing for the next
alright so let's do this
LH means i used l'hospital's rule
$\displaystyle \lim_{x\rightarrow 0}\frac{x\sin(6x)}{\sin^2(9x)}=LH=\lim_{x\rightarr ow 0}\frac{\sin(6x)+6x\cos(6x)}{2\cdot 9\sin(9x)\cos(9x)}$
Since we still get $\displaystyle \frac{0}{0}$ we use LH again
$\displaystyle =\lim_{x\rightarrow 0}\frac{6\cos(6x)+6\cos(6x)-36x\sin(6x)}{162\cos^2(9x)-162\sin^2(9x)}=\frac{12\cos(0)}{162\cos^2(0)}=\fra c{12}{162}=\frac{2}{27}$
any questions?
try the third one more time before I help you out there too