## anonymous 5 years ago Evaluate the following limit: lim x->0 sin(x) cos (x)/x

1. 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

2. anonymous

cos x/ x = 1?

3. myininaya

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

4. anonymous

i got DNE

5. anonymous

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

6. anonymous

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

7. myininaya

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

8. anonymous

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

9. anonymous

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

10. anonymous

no i wrote sinx / x * cosx / x

11. anonymous

it is not x^2 on the bottom

12. myininaya

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

13. 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?

14. anonymous

i seperate the fraction

15. myininaya

16. anonymous

sorry guys the lag go the best of me

17. anonymous

no i mean the 2nd line what you di

18. anonymous

did*

19. 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

20. anonymous

yes the 2nd line

21. anonymous

yes the 2nd line

22. myininaya

so we can write sinx/x*cosx?

23. anonymous

no sin x/x * cox x /x

24. myininaya

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

25. myininaya

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

26. myininaya

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

27. anonymous

so how do i work it? i am confused

28. myininaya

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

29. myininaya

as x gets closer to 0?*

30. anonymous

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

31. anonymous

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

32. myininaya

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

33. myininaya

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

34. anonymous

yes \\

35. myininaya

$\frac{5*6}{5}\neq \frac{5}{5}*\frac{6}{5}$

36. myininaya

do you agree with this statement?

37. anonymous

ok...

38. myininaya

$\frac{5*6}{5}=\frac{5}{5}*6=6$

39. myininaya

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

40. anonymous

k

41. myininaya

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

42. 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

43. myininaya

i gotta pee i might return

44. anonymous

lol ok THANK YOU SO MUCH