amyna
  • amyna
find the limit
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amyna
  • amyna
|dw:1444585400774:dw|
amyna
  • amyna
how do i do the l'hopitals rule for this problem?
freckles
  • freckles
I guess you are asking how to differentiate sin(pi*x) since I'm certain you know how to differentiate x

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amyna
  • amyna
so far i got: 1+cos pi x / 1-cos pi x ?
freckles
  • freckles
\[\frac{d}{dx}(\sin(\pi x)) =(\pi x)' \cos(\pi x) \text{ by chain rule }\]
freckles
  • freckles
well (pi*x)'=pi not 1
freckles
  • freckles
anyways you should be able to use direct sub after the first round of l'hospital once you differentiate sin(pi*x) correctly
amyna
  • amyna
so pi x + cos pi x / pi x - cos pi x ?
freckles
  • 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)
freckles
  • freckles
\[\frac{d}{dx}(x)=1 \\ \frac{d}{dx} \sin( \pi x)=(\pi x)' \cos(\pi x)= \pi \cos(\pi x)\]
amyna
  • amyna
oh okay so the answer is 1+ pi / 1- pi
freckles
  • freckles
yes that is right \[\lim_{x \rightarrow 0} \frac{x+ \sin( \pi x)}{x- \sin (\pi x)} \\ \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x + \sin(\pi x))}{\frac{d}{dx}(x- \sin(\pi x))} \\ \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x)+\frac{d}{dx}\sin(\pi x)}{\frac{d}{dx}(x)-\frac{d}{dx}\sin (\pi x)}\]
freckles
  • freckles
well almost it should be (1+pi)/(1-pi)
amistre64
  • 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?
amistre64
  • amistre64
i think my thought has an error tho
freckles
  • freckles
are you doing yours with l'hospital ? I was trying to follow all those broken line thingys...
amistre64
  • amistre64
i was trying to do long division :) to see what a series might be represented as
freckles
  • freckles
\[\lim_{x \rightarrow 0} \frac{x + \sin( \pi x)}{x- \sin( \pi x)} \\ \text{ divide top and bottom by } \pi x \\ \lim_{x \rightarrow 0} \frac{\frac{x}{\pi x}+\frac{\sin(\pi x)}{\pi x}}{\frac{x}{\pi x}-\frac{\sin(\pi x)}{\pi x}} \\ =\frac{\frac{1}{\pi}+1}{\frac{1}{\pi}-1} \\ \text{ now multiply top and bottom by } \pi \\ =\frac{1+\pi}{1-\pi}\]
amistre64
  • amistre64
squashing my creativity are you .... ;)
freckles
  • freckles
lol no let's try the long division thingy
freckles
  • freckles
or let me look at yours since I know what I'm looking at :p
amyna
  • amyna
lol Thanks for you help guys! :)
amistre64
  • 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.
freckles
  • freckles
your division seems great
freckles
  • freckles
so you get this : \[\lim_{x \rightarrow 0} (1+2 \frac{\sin(\pi x)}{x}+2 \cdot \frac{\sin^2( \pi x)}{x(x- \sin( \pi x)})\]
amistre64
  • 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.
freckles
  • freckles
\[\lim_{x \rightarrow 0} (1+2 \frac{\sin(\pi x)}{x}+2 \cdot \frac{\sin^2( \pi x)}{x(x- \sin( \pi x))})\]
amistre64
  • amistre64
i think i broke the wolf :/
1 Attachment
freckles
  • 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
freckles
  • freckles
I think I did something wrong though
freckles
  • freckles
oh no it was the same as previous answer

Looking for something else?

Not the answer you are looking for? Search for more explanations.