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.

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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
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@zarkon, et al, apparently she needs a non-l'hopital method, and i'll be darned if i can think of one
how about a power series solution :)
yes, do not pass go, do not collect $200

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Other answers:

lol
go directly to jail
in fact i think you need l'hopital twice at least
power series gives the quickest solution
\[\frac{\sec^2(x)-1}{3x^2}\] \[\frac{\tan^2(x)}{3x^2}\] etc
yes but if this is pre-derivative, then what can you do?
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
kidding :D
\[f(x)=\frac{sin x}{cosx}-x\] \[g(x)=x^3\] 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
I missed a cos x in the denominator but it won't make any difference as x-->0 cos x ---> 1
Zarkon what do you think , is my method correct?
the answer is 1/3
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)
Thanks I got your point

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