Can you help me find this limit: http://screencast.com/t/6CD9ErAG3N9m
All I need is the first step, I am gonna climb from there easily. Sorry if it's a little unclear picture, it's from an old book

- Christos

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- Christos

@ganeshie8 @Mertsj

- anonymous

for reference, the answer is \[-\sqrt3/2\] i do not remember the steps though, sorry

- Christos

I know the answer too my friend, but I dont know how to find it, thanks though

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## More answers

- anonymous

sorry i could not help :/

- anonymous

CANT EVEN SEE THE QUESTION

- Christos

How Come?? The guy above you saw it , I posted a picture

- Jhannybean

I can't see it either. what is the limit going to? Can't make it out after the arrow

- Christos

infinity+ ....

- Christos

+ infinity ***** I am sorry

- anonymous

lim of x going to positive infinity of sin((pi*x)/(2-3x))

- Jhannybean

\[\lim_{x \rightarrow +\infty}\sin (\frac{ \pi x }{ 2-3x })\]

- anonymous

can you actually take the pic clearly?

- rajee_sam

\[\lim_{x \rightarrow +\infty} Sin (\frac{ \pi x }{ 2-3x })\]

- Christos

Yes

- Jhannybean

Now Stewie can stop trolling, lol. xD

- anonymous

VICTORY?

- rajee_sam

I want to learn this too

- Jhannybean

Same here.

- anonymous

yes or no?

- rajee_sam

who is teaching??

- Jhannybean

You can divide by the highest power in the denominator which is x

- Christos

But how does that lead you to the solution which is -sqrt(3)/2

- Jhannybean

Can't you? \[\large \lim_{x \rightarrow +\infty}\sin(\frac{ \pi x/x }{ \frac{ 2 }{ x }-\frac{ 3x }{ x } })\]

- Christos

yes and then?

- Jhannybean

So what is \[\sin(-\frac{ \pi }{ 3 })\]? Whcih quadrant is sine negative ?

- rajee_sam

\[\lim_{x \rightarrow +\infty} \sin \frac{ x }{ x }(\frac{ \pi }{ \frac{ 2 }{ x } - 3 })\]

- Christos

@myko @timo86m @jim_thompson5910 @jhonyy9 @Hero @Euler271

- Christos

The result is most likely engative yea

- rajee_sam

Sin (- pi / 3)

- Christos

negative*

- Christos

yes but HOW @rajee_sam

- Jhannybean

Sine is negative in quadrant 3 and 4, so if \[\sin \frac{ \pi }{ 3 } = \frac{ \sqrt{3} }{ 2 }\] then \[\sin -\frac{ \pi }{ 3 }= -\frac{ \sqrt{3} }{ 2 }\]

- Christos

Am I eligible to differantiate this thing ? because if we differantiate we get sin(pi/-3)

- Christos

instantly

- rajee_sam

|dw:1369516434913:dw|

- Jhannybean

You just need to find the positive radian measure, which is \[\frac{ \pi }{ }\] and then translate it accordingly to find which negative value is correlated with the positive radian measure

- rajee_sam

\[\sin ( -\theta ) = - \sin (\theta)\]

- Jhannybean

sorry,i meant pi/3 *

- Christos

how did you get the x out @rajee_sam

- rajee_sam

I am taking a GCF for both numerator and denominator so that I can cancel them out and do not have any x multiplied with anything

- Christos

GCF ?

- rajee_sam

Common Factor

- phi

\[\lim_{x \rightarrow +\infty}\sin \left(\frac{ \pi x }{ 2-3x }\right) = \sin\left( \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x }\right)\]
as noted above you can rewrite
\[ \frac{ \pi x }{ 2-3x } = \frac{ \pi }{\frac{2}{x} -3} \]

- Christos

Alright! Alright! I solved it this way!
By during the time I was trying to solve this I figured out a second way!
Tell me if this is correct:
Just taking the derivative of both denominator and numerator and then we are done. Can I apply this to ANY limit? Because here it works!

- Jhannybean

that's what i did pi.... :(

- Christos

@phi

- Jhannybean

if you're taking the derivative you'll have to use chain rule on the whole thing, sin (whatever's inside) and then * whatever is inside.

- phi

yes, you can use L'Hopital's rule
http://en.wikipedia.org/wiki/L'HÃ´pital's_rule
if you get 0/0 or inf/inf

- Christos

and no chain rule??

- Christos

I am talking about this specific problem

- phi

no, use chain rule. use
\[ \lim_{x \rightarrow +\infty}\sin \left(\frac{ \pi x }{ 2-3x }\right) = \sin\left( \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x }\right) \]

- phi

*do not use chain rule

- Christos

but even if I dont use the chain rule we end up to the same result

- Christos

Ohh so this formula you showed to me just now is only when we have 0/0 or infinite/infinite ?

- Jhannybean

Oh I see, and you'd use LH rule to simplify it further and end up with the same result.

- phi

using L'Hopital's rule needs 0/0 or inf/inf
use it on
\[ \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x } \]

- Christos

ok and I get this http://screencast.com/t/jPlMQFrZ
What now?

- phi

you can use L'Hopital d top/ d bottom = pi/-3
or you can use algebra

- Christos

and the lim(x-->+infinity) goes away?

- Christos

oh I get you now

- Christos

but still I am stuck on the lim thingy

- phi

then sin (-pi/3) = -sqrt{3}/2
*** and the lim(x-->+infinity) goes away?
if you use L'Hopital
\[ \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x } = \lim_{x \rightarrow +\infty} \frac{\pi}{-3}= -\frac{\pi}{3}\]

- Christos

I see, thank you all

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