anonymous
  • anonymous
if f(x)=tanx-x and g(x)=x^3, evaluate the limit of f(x) over g(x) as x approaches 0. -3rd question.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@zarkon, et al, apparently she needs a non-l'hopital method, and i'll be darned if i can think of one
Zarkon
  • Zarkon
how about a power series solution :)
anonymous
  • anonymous
yes, do not pass go, do not collect $200

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Zarkon
  • Zarkon
lol
Zarkon
  • Zarkon
go directly to jail
anonymous
  • anonymous
in fact i think you need l'hopital twice at least
Zarkon
  • Zarkon
power series gives the quickest solution
anonymous
  • anonymous
\[\frac{\sec^2(x)-1}{3x^2}\] \[\frac{\tan^2(x)}{3x^2}\] etc
anonymous
  • anonymous
yes but if this is pre-derivative, then what can you do?
nenadmatematika
  • nenadmatematika
kaezalorene you can always say to your teacher that you've discovered some rule when you were home studying.....the rule that you can use the derivatives and solve....he will be suprised :D
nenadmatematika
  • nenadmatematika
kidding :D
ash2326
  • ash2326
\[f(x)=\frac{sin x}{cosx}-x\] \[g(x)=x^3\] we have \[f(x)/ g(x)=\frac{\frac{sin x}{cosx}-x}{x^3}\] dividing numerator and denominator by x now \[f(x)/g(x)= \frac{\frac{sin x}{x}-cos x}{x^2}\] \[limit x-->0 \frac{sinx}{x}=1\] so we have \[f(x)/g(x)= \frac{1-cos x}{x^2}\] 1-cos x = 2 (sin x/2)^2 so we get \[f(x)/g(x)= \frac{2 sin ^2 x/2}{x^2}\] dividing the numerator and denominator by 4 we get \[f(x)/g(x)=\frac {\frac{2 sin ^2 x/2}{4}}{\frac{x^2}{4}}\] \[(sin x / 2)/ (x/2-)-->1 as x---->0\] so we get \[f(x)/g(x)= \frac {2}{4}\] = 1/2
ash2326
  • ash2326
I missed a cos x in the denominator but it won't make any difference as x-->0 cos x ---> 1
ash2326
  • ash2326
Zarkon what do you think , is my method correct?
Zarkon
  • Zarkon
the answer is 1/3
Zarkon
  • Zarkon
unfortunatly you can't do it like that (the sin(x)/x is an integral part of the final solution...you can't takes its limit while holding everything else constant)
ash2326
  • ash2326
Thanks I got your point

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