## amyna one year ago find the limit

1. amyna

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2. amyna

how do i do the l'hopitals rule for this problem?

3. freckles

I guess you are asking how to differentiate sin(pi*x) since I'm certain you know how to differentiate x

4. amyna

so far i got: 1+cos pi x / 1-cos pi x ?

5. freckles

$\frac{d}{dx}(\sin(\pi x)) =(\pi x)' \cos(\pi x) \text{ by chain rule }$

6. freckles

well (pi*x)'=pi not 1

7. freckles

anyways you should be able to use direct sub after the first round of l'hospital once you differentiate sin(pi*x) correctly

8. amyna

so pi x + cos pi x / pi x - cos pi x ?

9. freckles

derivative of just x is 1... why do you put pi x? you still aren't writting the derivative of sin(pi*x) is pi*cos(pi*x)

10. freckles

$\frac{d}{dx}(x)=1 \\ \frac{d}{dx} \sin( \pi x)=(\pi x)' \cos(\pi x)= \pi \cos(\pi x)$

11. amyna

oh okay so the answer is 1+ pi / 1- pi

12. freckles

yes that is right $\lim_{x \rightarrow 0} \frac{x+ \sin( \pi x)}{x- \sin (\pi x)} \\ \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x + \sin(\pi x))}{\frac{d}{dx}(x- \sin(\pi x))} \\ \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x)+\frac{d}{dx}\sin(\pi x)}{\frac{d}{dx}(x)-\frac{d}{dx}\sin (\pi x)}$

13. freckles

well almost it should be (1+pi)/(1-pi)

14. amistre64

1+2sin(px)/x ------------ x-sin(px) | x + sin(px) (x -sin(px)) ------------ 2sin(px) (2sin(px) -2sin^2(px)/x) --------------------- 2sin^2(px)/x sin(px) = 0 when x=0, so i spose we could view this as 1+2sin(px)/x as x to 0, maybe?

15. amistre64

i think my thought has an error tho

16. freckles

are you doing yours with l'hospital ? I was trying to follow all those broken line thingys...

17. amistre64

i was trying to do long division :) to see what a series might be represented as

18. freckles

$\lim_{x \rightarrow 0} \frac{x + \sin( \pi x)}{x- \sin( \pi x)} \\ \text{ divide top and bottom by } \pi x \\ \lim_{x \rightarrow 0} \frac{\frac{x}{\pi x}+\frac{\sin(\pi x)}{\pi x}}{\frac{x}{\pi x}-\frac{\sin(\pi x)}{\pi x}} \\ =\frac{\frac{1}{\pi}+1}{\frac{1}{\pi}-1} \\ \text{ now multiply top and bottom by } \pi \\ =\frac{1+\pi}{1-\pi}$

19. amistre64

squashing my creativity are you .... ;)

20. freckles

lol no let's try the long division thingy

21. freckles

or let me look at yours since I know what I'm looking at :p

22. amyna

lol Thanks for you help guys! :)

23. amistre64

maybe fliiping the order of the terms might help .. one way is x to infinity, the other is x to zero ... ive noticed this but i cant see why.

24. freckles

your division seems great

25. freckles

so you get this : $\lim_{x \rightarrow 0} (1+2 \frac{\sin(\pi x)}{x}+2 \cdot \frac{\sin^2( \pi x)}{x(x- \sin( \pi x)})$

26. amistre64

i was thinking we could factor our a sin(px) to zero out all but the first 2 terms .. but that seems to be where my idea goes awry.

27. freckles

$\lim_{x \rightarrow 0} (1+2 \frac{\sin(\pi x)}{x}+2 \cdot \frac{\sin^2( \pi x)}{x(x- \sin( \pi x))})$

28. amistre64

i think i broke the wolf :/

29. freckles

$1+2 \pi +2 \cdot \lim_{x \rightarrow 0} \frac{\sin^2(\pi x)}{x^2- x \sin(\pi x)} \\ \\ \text{ for the last quotient there divide top and bottom by } x^2 \\ 1+2 \pi +2 \cdot \frac{\pi^2}{1-\pi}$ I still go back to my old ways using that one famous limit thingy

30. freckles

I think I did something wrong though

31. freckles

oh no it was the same as previous answer