Question 927647: NOTE: THIS IS A CALCULUS INDETERMINATE FORM, L'HOSPITAL'S RULE PROBLEM. Any help is greatly appreciated!
Find the limit as x approaches 0 when sin 8x/ tan 9x
I know the answer is 8/9, but I don't understand why. This is a L'Hopital rule problem, so I tried that. When I took the derivative of the top, over the derivative of the bottom I got 8 cos 8x/ 9 (sec 9x)^2. When you plug in 0 into this, you get 8/(9/0), so this would simplify to be 0/9. If a limit approaches 0/9, that means that is is a very small number being divided by a constant, so it would approach negative infinity. But that is not what it turns out to be.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! The mistake you made was thinking that  when it is really  since 
---------------------------------------------------------------------------
Let
 = \sin(8x) )
 = \tan(9x) )
As x approaches 0, f(x) approaches 0. This is because
 = f(0) = \sin(8*0) = \sin(0) = 0 )
(this limit law works because sine is a continuous function).
For similar reasons, as x approaches 0, g(x) approaches 0 (mainly because tan(x) = sin(x)/cos(x))
So as  , then }{g(x)} \to \frac{0}{0}) which is indeterminate. You are correct about that.
Because }{g(x)} = \frac{0}{0}) , which is indeterminate, we'll have to use L'Hopital's Rule (or L'Hospital's Rule) to see if the limit exists. It may or may not.
=======================================================================
Applying L'Hopital's Rule (or L'Hospital's Rule)
 = 8\cos(8x) )
-------------------------------------------------------
 = 9\sec^2(9x) )
-------------------------------------------------------
}{g(x)} = \lim_{x\to0}\frac{f'(x)}{g'(x)} \ ... \text{LHospitals rule})
}{\tan(9x)} = \lim_{x\to0}\frac{8\cos(8x)}{9\sec^2(9x)})
}{\tan(9x)} = \lim_{x\to0}\frac{8\cos(8x)}{9*\frac{1}{cos^2(9x)}})
}{\tan(9x)} = \frac{8\cos(8*0)}{9*\frac{1}{cos^2(9*0)}})
}{\tan(9x)} = \frac{8\cos(0)}{9*\frac{1}{cos^2(0)}})
}{\tan(9x)} = \frac{8*1}{9*\frac{1}{1^2}})
}{\tan(9x)} = \frac{8*1}{9*\frac{1}{1}})
}{\tan(9x)} = \frac{8*1}{9*1})
}{\tan(9x)} = \frac{8}{9})
------------------------------------------------------------------------------------------------------------------------
If you need more one-on-one help, email me at jim_thompson5910@hotmail.com. You can ask me a few more questions for free, but afterwards, I would charge you ($2 a problem to have steps shown or $1 a problem for answer only).
Alternatively, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Any amount is greatly appreciated as it helps me a lot. This donation is to support free tutoring. Thank you.
Jim
------------------------------------------------------------------------------------------------------------------------
|
|
|
