Evaluate the following limit:
lim x->0 sin(x) cos (x)/x

- anonymous

Evaluate the following limit:
lim x->0 sin(x) cos (x)/x

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- anonymous

\[\lim_{x \rightarrow 0}sinxcosx/x=\lim_{x \rightarrow 0}sinx/x*\lim_{x \rightarrow 0}\cos(x)\]
both limits are 1. So the whole limit is 1

- anonymous

cos x/ x = 1?

- myininaya

sinx/x->1
cosx is left over but also goes to 1
both as x->0

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## More answers

- anonymous

i got DNE

- anonymous

no the limit exists, it is 1. cos(x) goes to 1 and sinx/x goes to 1 as well

- anonymous

because cos x / x = 1/0... no?

- myininaya

lol can you not write the function as sinx/x *cosx?

- anonymous

you aren't plugging in 0 though it's a limit

- anonymous

there are limits that have a division by 0 and still exist it is about infinitesimal behavior

- anonymous

no i wrote sinx / x * cosx / x

- anonymous

it is not x^2 on the bottom

- myininaya

(5*6)/(5)=6
but by what you say we can do or suppose to do
5/5*6/5=6/5

- myininaya

i mean (5*6)/5=5/5*6=1*6=6
and you say (5*6)/5=5/5*6/5=6/5
is that what you are saying this second line right here?

- anonymous

i seperate the fraction

- myininaya

i think you are thinking about addition on fractions
(a+b)/c=a/c+b/c

- anonymous

sorry guys the lag go the best of me

- anonymous

no i mean the 2nd line what you di

- anonymous

did*

- myininaya

you can't do (5*6)/5=5/5*6/5
but you do (5*6)/5=5/5*6 or 5/6*5=6

- anonymous

yes the 2nd line

- anonymous

yes the 2nd line

- myininaya

so we can write sinx/x*cosx?

- anonymous

no sin x/x * cox x /x

- myininaya

but you just said we can't do (5*6)/5=5/5*6/5

- myininaya

(sinx*cosx)/x=sinx/x * cosx

- myininaya

or =cosx/x * sinx ( it does us no good to write it like this though)

- anonymous

so how do i work it? i am confused

- myininaya

write (sinx*cosx)/x=sinx/x*cosx
where does sinx/x go to as x gets closer to 1?

- myininaya

as x gets closer to 0?*

- anonymous

do you know how to use latex here because i dont understand what you typed

- anonymous

so the answer is 1 i dont undertstand why it cant be done the way i did it i understand what you typed

- myininaya

so you still don't understand why we can't do:
\[\frac{\sin(x)\cos(x)}{x}=\frac{\sin(x)}{x}\frac{\cos(x)}{x}?\]

- myininaya

we don't have x^2 on bottom? we only have x

- anonymous

yes
\\

- myininaya

\[\frac{5*6}{5}\neq \frac{5}{5}*\frac{6}{5}\]

- myininaya

do you agree with this statement?

- anonymous

ok...

- myininaya

\[\frac{5*6}{5}=\frac{5}{5}*6=6\]

- myininaya

\[\lim_{x \rightarrow 0} \frac{\sin(x)\cos(x)}{x}=\lim_{x \rightarrow 0}\frac{\sin(x)}{x}*\cos(x)\]

- anonymous

k

- myininaya

\[=\lim_{x \rightarrow 0}\frac{\sin(x)}{x}*\lim_{x \rightarrow 0}\cos(x)=1*1=1\]

- anonymous

i guess i gotta recheck my fractions thank you so much i have 1 more question if your not busy can you help me with that please

- myininaya

i gotta pee i might return

- anonymous

lol ok THANK YOU SO MUCH

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