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)
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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)
do you know l'hospital's rule?
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
Quote:
Originally Posted by katekate
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
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
Quote:
Originally Posted by katekate
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
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
