## anonymous one year ago Trigonometric limit help.

1. anonymous

2. anonymous

I can't figure out how to fix the denominator in any way.

3. freckles

$\lim_{x \rightarrow 0} \frac{x^3 \sin(2x)}{\cos(x^4-\frac{\pi}{2}) \tan(x^3+x)}$ writing it here so I don't have to keep looking back and forth and now thinking...

4. anonymous

Yep no problems haha.

5. anonymous

Take a look at my work too. I figured out almost all of it.

6. freckles

what happens when you plug in 0 we should get 0/0 so we definitely have more work to do...

7. anonymous

It's just that pesky first limit.

8. anonymous

If I was allowed to use taylor polynomials this would be a joke but I am not.

9. freckles

yeah I was looking at your work and that one term we have a problem with since that first limit doesn't actually exist $\lim_{x \rightarrow 0} \frac{x^3\cos(x^3+x)(2x)}{x^4(x^3+x)} \\ \lim_{x \rightarrow 0} 2 \frac{\cos(x^3+x)}{x^3+x} \\ \text{ plug \in 0 you get } \frac{1}{0} \\ \text{ so the limit does not exist } \\ \text{ so I don't think we can take the limit of each factor }$

10. freckles

you get 2/0 sorry forgot to multiply the 2 on top to the 1 on top

11. freckles

still doesn't exist though

12. anonymous

According to wolfram, it's 4 though.

13. freckles

for the one I just said doesn't exist?

14. anonymous

WAIT.

15. anonymous

OMG> Sorry correction. The sin(2x) in the numerator should be a sin^2(2x) .

16. freckles

so the work on that one page is slightly off then because of this new factor sin(2x)

17. anonymous

Yep let me try again...

18. freckles

$\lim_{x \rightarrow 0} \frac{ x^3 \cos(x^3+x) (2x) \sin(2x)}{x^4(x^3+x)} \\ \lim_{x \rightarrow 0} \frac{\sin(2x)}{2x} (2) \frac{x^3 \cos(x^3+x)(2x)}{x^3(x^3+x) } \\ \lim_{x \rightarrow 0} 2 \frac{\cos(x^3+x)}{x^3+x}(2x) \\ \lim_{x \rightarrow 0 } 4 \frac{\cos(x^3+x)}{x^2+1}$ plug in 0 :)

19. freckles

and dido awesome algebra going on your work

20. anonymous

Omg thanks T_T . I can't believe a typo threw me off.

21. anonymous

Haha thanks!

22. freckles

yep throwing in that sin(2x) into your first factor there totally fixed it so the limit would exist