anonymous
  • anonymous
Evaluate the following limit: lim x->0 sin(x) cos (x)/x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
cos x/ x = 1?
myininaya
  • myininaya
sinx/x->1 cosx is left over but also goes to 1 both as x->0

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anonymous
  • anonymous
i got DNE
anonymous
  • anonymous
no the limit exists, it is 1. cos(x) goes to 1 and sinx/x goes to 1 as well
anonymous
  • anonymous
because cos x / x = 1/0... no?
myininaya
  • myininaya
lol can you not write the function as sinx/x *cosx?
anonymous
  • anonymous
you aren't plugging in 0 though it's a limit
anonymous
  • anonymous
there are limits that have a division by 0 and still exist it is about infinitesimal behavior
anonymous
  • anonymous
no i wrote sinx / x * cosx / x
anonymous
  • 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?
anonymous
  • anonymous
i seperate the fraction
myininaya
  • myininaya
i think you are thinking about addition on fractions (a+b)/c=a/c+b/c
anonymous
  • anonymous
sorry guys the lag go the best of me
anonymous
  • anonymous
no i mean the 2nd line what you di
anonymous
  • 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
anonymous
  • anonymous
yes the 2nd line
anonymous
  • anonymous
yes the 2nd line
myininaya
  • myininaya
so we can write sinx/x*cosx?
anonymous
  • 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)
anonymous
  • 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?*
anonymous
  • anonymous
do you know how to use latex here because i dont understand what you typed
anonymous
  • 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
anonymous
  • anonymous
yes \\
myininaya
  • myininaya
\[\frac{5*6}{5}\neq \frac{5}{5}*\frac{6}{5}\]
myininaya
  • myininaya
do you agree with this statement?
anonymous
  • anonymous
ok...
myininaya
  • myininaya
\[\frac{5*6}{5}=\frac{5}{5}*6=6\]
myininaya
  • myininaya
\[\lim_{x \rightarrow 0} \frac{\sin(x)\cos(x)}{x}=\lim_{x \rightarrow 0}\frac{\sin(x)}{x}*\cos(x)\]
anonymous
  • anonymous
k
myininaya
  • myininaya
\[=\lim_{x \rightarrow 0}\frac{\sin(x)}{x}*\lim_{x \rightarrow 0}\cos(x)=1*1=1\]
anonymous
  • 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
anonymous
  • anonymous
lol ok THANK YOU SO MUCH

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