document.write( "Question 927647: NOTE: THIS IS A CALCULUS INDETERMINATE FORM, L'HOSPITAL'S RULE PROBLEM. Any help is greatly appreciated!\r
\n" ); document.write( "\n" ); document.write( "Find the limit as x approaches 0 when sin 8x/ tan 9x
\n" ); document.write( "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.
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Algebra.Com's Answer #563132 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
The mistake you made was thinking that \"sec%280%29+=+0\" when it is really \"sec%280%29+=+1\" since \"sec%28x%29+=+1%2Fcos%28x%29\"\r
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\n" ); document.write( "\n" ); document.write( "Let \r
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\n" ); document.write( "\n" ); document.write( "As x approaches 0, f(x) approaches 0. This is because
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\n" ); document.write( "(this limit law works because sine is a continuous function).\r
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\n" ); document.write( "\n" ); document.write( "For similar reasons, as x approaches 0, g(x) approaches 0 (mainly because tan(x) = sin(x)/cos(x))\r
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\n" ); document.write( "\n" ); document.write( "So as , then which is indeterminate. You are correct about that.\r
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\n" ); document.write( "\n" ); document.write( "Because , 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.\r
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\n" ); document.write( "\n" ); document.write( "Applying L'Hopital's Rule (or L'Hospital's Rule)\r
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\n" ); document.write( "\n" ); document.write( "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).\r
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