## anonymous 3 years ago Find the limit. lim 3 − 3 tan x/sin x − cos x x→π/4

1. anonymous

Could you write out how this looks. the division sign doesn't really indicate where the equation is divided.

2. anonymous

At least put parenthesis where they belong.

3. anonymous

lim (3 − 3 tan x)/(sin x − cos x) x→π/4

4. anonymous

First try to just put in the value pi/4 and decide if the function is defined in that point

5. anonymous

you'll get -3sqrt(2)

6. anonymous

You see that sin pi/4- cos pi/4 equals zero so the function is undefined at pi/4 so what you want to do is to simplify the function

7. anonymous

okaaaaaaaaay thambi

8. anonymous

how do i simplify this?

9. anonymous

you can turn your denominator into$(3-3(\frac{ cosx }{ sinx })$ then split the numerator into two different fractions$\frac{ 3 }{ sinx-cosx } - \frac{ \frac{ sinx }{ cosx } }{ sinx-cosx }$

10. anonymous

sorry meant numerator

11. anonymous

and that should have been three times sine over cosine sorry I am messing it up pretty bad

12. anonymous

loll itss okayy

13. anonymous

i dont get the question

14. anonymous

$\frac{ 3-3\tan(x) }{ \sin(x) - \cos(x) } \times \frac{ \sin(x) +\cos(x) }{ \sin(x) +\cos(x) }$

15. anonymous

$\frac{ 3\sin(x) - 3\frac{ \sin^2(x) }{ \cos(x) } + 3\cos(x) -3\sin(x) }{ \sin^2(x) -\cos^2(x) }$

16. anonymous

$\frac{ - 3\frac{ \sin^2(x) }{ \cos(x) } + 3\cos(x) }{ \sin^2(x) -\cos^2(x) }$

17. anonymous

$\frac{ - 3\sin^2(x) + 3\cos^2(x) }{ \cos(x)[\sin^2(x) -\cos^2(x)] } = \frac{ -3 }{ \cos(x) }$

18. anonymous

sorry i was busy

19. anonymous

so i sub in pi/4

20. anonymous

yes

21. anonymous

-3/(1/sqroot2)

22. anonymous

which simplifies into -3sqrt(2)

23. anonymous

-sqroot2/3

24. anonymous

no .. -3 * sqrt(2)

25. anonymous

yeye

26. anonymous

i gett it

27. anonymous

$-3\sqrt2$

28. anonymous

like this

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