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
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Christos
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@ganeshie8 @Mertsj
robz8
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for reference, the answer is \[-\sqrt3/2\] i do not remember the steps though, sorry
Christos
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I know the answer too my friend, but I dont know how to find it, thanks though
robz8
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sorry i could not help :/
Stewie_Griffin_
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CANT EVEN SEE THE QUESTION
Christos
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How Come?? The guy above you saw it , I posted a picture
Jhannybean
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I can't see it either. what is the limit going to? Can't make it out after the arrow
Christos
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infinity+ ....
Christos
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+ infinity ***** I am sorry
robz8
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lim of x going to positive infinity of sin((pi*x)/(2-3x))
Jhannybean
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\[\lim_{x \rightarrow +\infty}\sin (\frac{ \pi x }{ 2-3x })\]
Stewie_Griffin_
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can you actually take the pic clearly?
rajee_sam
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\[\lim_{x \rightarrow +\infty} Sin (\frac{ \pi x }{ 2-3x })\]
Christos
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Yes
Jhannybean
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Now Stewie can stop trolling, lol. xD
Stewie_Griffin_
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VICTORY?
rajee_sam
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I want to learn this too
Jhannybean
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Same here.
Stewie_Griffin_
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yes or no?
rajee_sam
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who is teaching??
Jhannybean
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You can divide by the highest power in the denominator which is x
Christos
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But how does that lead you to the solution which is -sqrt(3)/2
Jhannybean
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Can't you? \[\large \lim_{x \rightarrow +\infty}\sin(\frac{ \pi x/x }{ \frac{ 2 }{ x }-\frac{ 3x }{ x } })\]
Christos
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yes and then?
Jhannybean
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So what is \[\sin(-\frac{ \pi }{ 3 })\]? Whcih quadrant is sine negative ?
rajee_sam
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\[\lim_{x \rightarrow +\infty} \sin \frac{ x }{ x }(\frac{ \pi }{ \frac{ 2 }{ x } - 3 })\]
Christos
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@myko @timo86m @jim_thompson5910 @jhonyy9 @Hero @Euler271
Christos
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The result is most likely engative yea
rajee_sam
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Sin (- pi / 3)
Christos
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negative*
Christos
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yes but HOW @rajee_sam
Jhannybean
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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
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Am I eligible to differantiate this thing ? because if we differantiate we get sin(pi/-3)
Christos
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instantly
rajee_sam
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|dw:1369516434913:dw|
Jhannybean
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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
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\[\sin ( -\theta ) = - \sin (\theta)\]
Jhannybean
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sorry,i meant pi/3 *
Christos
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how did you get the x out @rajee_sam
rajee_sam
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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
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GCF ?
rajee_sam
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Common Factor
phi
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\[\lim_{x \rightarrow +\infty}\sin \left(\frac{ \pi x }{ 2-3x }\right) = \sin\left( \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x }\right)\]
as noted above you can rewrite
\[ \frac{ \pi x }{ 2-3x } = \frac{ \pi }{\frac{2}{x} -3} \]
Christos
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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
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that's what i did pi.... :(
Christos
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@phi
Jhannybean
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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.
Christos
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and no chain rule??
Christos
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I am talking about this specific problem
phi
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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
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*do not use chain rule
Christos
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but even if I dont use the chain rule we end up to the same result
Christos
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Ohh so this formula you showed to me just now is only when we have 0/0 or infinite/infinite ?
Jhannybean
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Oh I see, and you'd use LH rule to simplify it further and end up with the same result.
phi
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using L'Hopital's rule needs 0/0 or inf/inf
use it on
\[ \lim_{x \rightarrow +\infty} \frac{ \pi x }{ 2-3x } \]
phi
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you can use L'Hopital d top/ d bottom = pi/-3
or you can use algebra
Christos
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and the lim(x-->+infinity) goes away?
Christos
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oh I get you now
Christos
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but still I am stuck on the lim thingy
phi
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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
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I see, thank you all