Question 927647
The mistake you made was thinking that {{{sec(0) = 0}}} when it is really {{{sec(0) = 1}}} since {{{sec(x) = 1/cos(x)}}}


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Let 

*[Tex \LARGE f(x) = \sin(8x) ]
*[Tex \LARGE g(x) = \tan(9x) ]


As x approaches 0, f(x) approaches 0. This is because 
*[Tex \LARGE \lim_{x\to 0}f(x) = 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 *[Tex \LARGE x \to 0], then *[Tex \LARGE \frac{f(x)}{g(x)} \to \frac{0}{0}] which is indeterminate. You are correct about that.


Because *[Tex \LARGE \lim_{x\to0}\frac{f(x)}{g(x)} = \frac{0}{0}], which is indeterminate, we'll have to use <a href="http://mathworld.wolfram.com/LHospitalsRule.html">L'Hopital's Rule (or L'Hospital's Rule)</a> 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)


*[Tex \LARGE f(x) = \sin(8x) ]


*[Tex \LARGE f'(x) = 8\cos(8x) ]

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*[Tex \LARGE g(x) = \tan(9x) ]


*[Tex \LARGE g'(x) = 9\sec^2(9x) ]

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*[Tex \LARGE \lim_{x\to0}\frac{f(x)}{g(x)} = \lim_{x\to0}\frac{f'(x)}{g'(x)} \ ... \text{LHospitals rule}]



*[Tex \LARGE \lim_{x\to0}\frac{\sin(8x)}{\tan(9x)} = \lim_{x\to0}\frac{8\cos(8x)}{9\sec^2(9x)}]



*[Tex \LARGE \lim_{x\to0}\frac{\sin(8x)}{\tan(9x)} = \lim_{x\to0}\frac{8\cos(8x)}{9*\frac{1}{cos^2(9x)}}]



*[Tex \LARGE \lim_{x\to0}\frac{\sin(8x)}{\tan(9x)} = \frac{8\cos(8*0)}{9*\frac{1}{cos^2(9*0)}}]



*[Tex \LARGE \lim_{x\to0}\frac{\sin(8x)}{\tan(9x)} = \frac{8\cos(0)}{9*\frac{1}{cos^2(0)}}]



*[Tex \LARGE \lim_{x\to0}\frac{\sin(8x)}{\tan(9x)} = \frac{8*1}{9*\frac{1}{1^2}}]



*[Tex \LARGE \lim_{x\to0}\frac{\sin(8x)}{\tan(9x)} = \frac{8*1}{9*\frac{1}{1}}]



*[Tex \LARGE \lim_{x\to0}\frac{\sin(8x)}{\tan(9x)} = \frac{8*1}{9*1}]



*[Tex \LARGE \lim_{x\to0}\frac{\sin(8x)}{\tan(9x)} = \frac{8}{9}]

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