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

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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

Mathematics
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for reference, the answer is \[-\sqrt3/2\] i do not remember the steps though, sorry
I know the answer too my friend, but I dont know how to find it, thanks though

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Other answers:

sorry i could not help :/
CANT EVEN SEE THE QUESTION
How Come?? The guy above you saw it , I posted a picture
I can't see it either. what is the limit going to? Can't make it out after the arrow
infinity+ ....
+ infinity ***** I am sorry
lim of x going to positive infinity of sin((pi*x)/(2-3x))
\[\lim_{x \rightarrow +\infty}\sin (\frac{ \pi x }{ 2-3x })\]
can you actually take the pic clearly?
\[\lim_{x \rightarrow +\infty} Sin (\frac{ \pi x }{ 2-3x })\]
Yes
Now Stewie can stop trolling, lol. xD
VICTORY?
I want to learn this too
Same here.
yes or no?
who is teaching??
You can divide by the highest power in the denominator which is x
But how does that lead you to the solution which is -sqrt(3)/2
Can't you? \[\large \lim_{x \rightarrow +\infty}\sin(\frac{ \pi x/x }{ \frac{ 2 }{ x }-\frac{ 3x }{ x } })\]
yes and then?
So what is \[\sin(-\frac{ \pi }{ 3 })\]? Whcih quadrant is sine negative ?
\[\lim_{x \rightarrow +\infty} \sin \frac{ x }{ x }(\frac{ \pi }{ \frac{ 2 }{ x } - 3 })\]
The result is most likely engative yea
Sin (- pi / 3)
negative*
yes but HOW @rajee_sam
Sine is negative in quadrant 3 and 4, so if \[\sin \frac{ \pi }{ 3 } = \frac{ \sqrt{3} }{ 2 }\] then \[\sin -\frac{ \pi }{ 3 }= -\frac{ \sqrt{3} }{ 2 }\]
Am I eligible to differantiate this thing ? because if we differantiate we get sin(pi/-3)
instantly
|dw:1369516434913:dw|
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
\[\sin ( -\theta ) = - \sin (\theta)\]
sorry,i meant pi/3 *
how did you get the x out @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
GCF ?
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 \[ \frac{ \pi x }{ 2-3x } = \frac{ \pi }{\frac{2}{x} -3} \]
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!
that's what i did pi.... :(
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'Hôpital's_rule if you get 0/0 or inf/inf
and no chain rule??
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
but even if I dont use the chain rule we end up to the same result
Ohh so this formula you showed to me just now is only when we have 0/0 or infinite/infinite ?
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 \[ \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x } \]
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
and the lim(x-->+infinity) goes away?
oh I get you now
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 \[ \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x } = \lim_{x \rightarrow +\infty} \frac{\pi}{-3}= -\frac{\pi}{3}\]
I see, thank you all

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