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anonymous
 4 years ago
if f(x)=tanxx and g(x)=x^3, evaluate the limit of f(x) over g(x) as x approaches 0. 3rd question.
anonymous
 4 years ago
if f(x)=tanxx and g(x)=x^3, evaluate the limit of f(x) over g(x) as x approaches 0. 3rd question.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@zarkon, et al, apparently she needs a nonl'hopital method, and i'll be darned if i can think of one

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0how about a power series solution :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, do not pass go, do not collect $200

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in fact i think you need l'hopital twice at least

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0power series gives the quickest solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\sec^2(x)1}{3x^2}\] \[\frac{\tan^2(x)}{3x^2}\] etc

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes but if this is prederivative, then what can you do?

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.0kaezalorene 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

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\frac{sin x}{cosx}x\] \[g(x)=x^3\] 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 \[f(x)/g(x)= \frac{1cos x}{x^2}\] 1cos 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
 4 years ago
Best ResponseYou've already chosen the best response.0I missed a cos x in the denominator but it won't make any difference as x>0 cos x > 1

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0Zarkon what do you think , is my method correct?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0unfortunatly 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
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks I got your point
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