Christos
  • Christos
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Christos
  • Christos
@ganeshie8 @Mertsj
anonymous
  • anonymous
for reference, the answer is \[-\sqrt3/2\] i do not remember the steps though, sorry
Christos
  • Christos
I know the answer too my friend, but I dont know how to find it, thanks though

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anonymous
  • anonymous
sorry i could not help :/
anonymous
  • anonymous
CANT EVEN SEE THE QUESTION
Christos
  • Christos
How Come?? The guy above you saw it , I posted a picture
Jhannybean
  • Jhannybean
I can't see it either. what is the limit going to? Can't make it out after the arrow
Christos
  • Christos
infinity+ ....
Christos
  • Christos
+ infinity ***** I am sorry
anonymous
  • anonymous
lim of x going to positive infinity of sin((pi*x)/(2-3x))
Jhannybean
  • Jhannybean
\[\lim_{x \rightarrow +\infty}\sin (\frac{ \pi x }{ 2-3x })\]
anonymous
  • anonymous
can you actually take the pic clearly?
rajee_sam
  • rajee_sam
\[\lim_{x \rightarrow +\infty} Sin (\frac{ \pi x }{ 2-3x })\]
Christos
  • Christos
Yes
Jhannybean
  • Jhannybean
Now Stewie can stop trolling, lol. xD
anonymous
  • anonymous
VICTORY?
rajee_sam
  • rajee_sam
I want to learn this too
Jhannybean
  • Jhannybean
Same here.
anonymous
  • anonymous
yes or no?
rajee_sam
  • rajee_sam
who is teaching??
Jhannybean
  • Jhannybean
You can divide by the highest power in the denominator which is x
Christos
  • Christos
But how does that lead you to the solution which is -sqrt(3)/2
Jhannybean
  • Jhannybean
Can't you? \[\large \lim_{x \rightarrow +\infty}\sin(\frac{ \pi x/x }{ \frac{ 2 }{ x }-\frac{ 3x }{ x } })\]
Christos
  • Christos
yes and then?
Jhannybean
  • Jhannybean
So what is \[\sin(-\frac{ \pi }{ 3 })\]? Whcih quadrant is sine negative ?
rajee_sam
  • rajee_sam
\[\lim_{x \rightarrow +\infty} \sin \frac{ x }{ x }(\frac{ \pi }{ \frac{ 2 }{ x } - 3 })\]
Christos
  • Christos
@myko @timo86m @jim_thompson5910 @jhonyy9 @Hero @Euler271
Christos
  • Christos
The result is most likely engative yea
rajee_sam
  • rajee_sam
Sin (- pi / 3)
Christos
  • Christos
negative*
Christos
  • Christos
yes but HOW @rajee_sam
Jhannybean
  • 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
  • Christos
Am I eligible to differantiate this thing ? because if we differantiate we get sin(pi/-3)
Christos
  • Christos
instantly
rajee_sam
  • rajee_sam
|dw:1369516434913:dw|
Jhannybean
  • 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
  • rajee_sam
\[\sin ( -\theta ) = - \sin (\theta)\]
Jhannybean
  • Jhannybean
sorry,i meant pi/3 *
Christos
  • Christos
how did you get the x out @rajee_sam
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
  • Christos
GCF ?
rajee_sam
  • rajee_sam
Common Factor
phi
  • 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
  • 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
  • Jhannybean
that's what i did pi.... :(
Christos
  • Christos
@phi
Jhannybean
  • 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
  • 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
Christos
  • Christos
and no chain rule??
Christos
  • Christos
I am talking about this specific problem
phi
  • 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
  • phi
*do not use chain rule
Christos
  • Christos
but even if I dont use the chain rule we end up to the same result
Christos
  • Christos
Ohh so this formula you showed to me just now is only when we have 0/0 or infinite/infinite ?
Jhannybean
  • Jhannybean
Oh I see, and you'd use LH rule to simplify it further and end up with the same result.
phi
  • 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
  • Christos
ok and I get this http://screencast.com/t/jPlMQFrZ What now?
phi
  • phi
you can use L'Hopital d top/ d bottom = pi/-3 or you can use algebra
Christos
  • Christos
and the lim(x-->+infinity) goes away?
Christos
  • Christos
oh I get you now
Christos
  • Christos
but still I am stuck on the lim thingy
phi
  • 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
  • Christos
I see, thank you all

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