SOLUTION: 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
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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
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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.
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Applying L'Hopital's Rule (or L'Hospital's Rule)
 = 8\cos(8x) )
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 = 9\sec^2(9x) )
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}{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})
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