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cos x/ x = 1?

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

i got DNE

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

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

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

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

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

no i wrote sinx / x * cosx / x

it is not x^2 on the bottom

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

i seperate the fraction

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

sorry guys the lag go the best of me

no i mean the 2nd line what you di

did*

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

yes the 2nd line

yes the 2nd line

so we can write sinx/x*cosx?

no sin x/x * cox x /x

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

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

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

so how do i work it? i am confused

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

as x gets closer to 0?*

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

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

yes
\\

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

do you agree with this statement?

ok...

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

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

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

i gotta pee i might return

lol ok THANK YOU SO MUCH