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Mathematics 15 Online
OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

cos x/ x = 1?

myininaya (myininaya):

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

OpenStudy (anonymous):

i got DNE

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

no i wrote sinx / x * cosx / x

OpenStudy (anonymous):

it is not x^2 on the bottom

myininaya (myininaya):

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

myininaya (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?

OpenStudy (anonymous):

i seperate the fraction

myininaya (myininaya):

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

OpenStudy (anonymous):

sorry guys the lag go the best of me

OpenStudy (anonymous):

no i mean the 2nd line what you di

OpenStudy (anonymous):

did*

myininaya (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

OpenStudy (anonymous):

yes the 2nd line

OpenStudy (anonymous):

yes the 2nd line

myininaya (myininaya):

so we can write sinx/x*cosx?

OpenStudy (anonymous):

no sin x/x * cox x /x

myininaya (myininaya):

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

myininaya (myininaya):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

so how do i work it? i am confused

myininaya (myininaya):

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

myininaya (myininaya):

as x gets closer to 0?*

OpenStudy (anonymous):

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

OpenStudy (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 (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 (myininaya):

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

OpenStudy (anonymous):

yes \\

myininaya (myininaya):

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

myininaya (myininaya):

do you agree with this statement?

OpenStudy (anonymous):

ok...

myininaya (myininaya):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

k

myininaya (myininaya):

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

OpenStudy (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 (myininaya):

i gotta pee i might return

OpenStudy (anonymous):

lol ok THANK YOU SO MUCH

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