Question

lim x->0, (sin 9x)/(x cos x),,, can someone help?

Answer 1

9

Answer 2

write

\[\frac{\sin(9x)}{x}\]

as

\[9\frac{\sin(9x)}{9x}\]
\[\frac{\sin(9x)}{9x}\to 1\text{ as }x\to 0\]

Answer 3

where does the cos x go?

Answer 4

\[\cos(x)\to 1\text{ as }x\to 0\]

Answer 5

\[\cos(0)=1\]

Answer 6

i didnt get it,, u substitute x with 0 is it?

Answer 7

\[\lim_{x \rightarrow 0}\frac{\sin(9x)}{x}\times \lim_{x\rightarrow 0}\frac{1}{\cos(x)}\]
\[\lim_{x \rightarrow 0}\frac{9\sin(9x)}{9x}\times \lim_{x\rightarrow 0}\frac{1}{\cos(x)}\]
\[9\lim_{x \rightarrow 0}\frac{\sin(9x)}{9x}\times \lim_{x\rightarrow 0}\frac{1}{\cos(x)}\]
first and second limit is 1 the first because you know
\[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1\] the second by replacing x by 1 (since it is continuous no problem doing that)

Answer 8

yes! tanx :D

Answer 9

:)

Answer 10

MEDAL PLSS

Answer 11

medal for wht? u dont even answer my ques :D

Answer 12

You could also use L'Hopitals.
\[\lim_{x\rightarrow0} \frac{9\cos(9x)}{\cos(x)-x\sin(x)}\]