Question

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.

Answer 1

@zarkon, et al, apparently she needs a non-l'hopital method, and i'll be darned if i can think of one

Answer 2

how about a power series solution :)

Answer 3

yes, do not pass go, do not collect $200

Answer 4

lol

Answer 5

go directly to jail

Answer 6

in fact i think you need l'hopital twice at least

Answer 7

power series gives the quickest solution

Answer 8

\[\frac{\sec^2(x)-1}{3x^2}\]
\[\frac{\tan^2(x)}{3x^2}\] etc

Answer 9

yes but if this is pre-derivative, then what can you do?

Answer 10

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

Answer 11

kidding :D

Answer 12

\[f(x)=\frac{sin x}{cosx}-x\]
\[g(x)=x^3\]
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

Answer 13

I missed a cos x in the denominator but it won't make any difference as x-->0 cos x ---> 1

Answer 14

Zarkon what do you think , is my method correct?

Answer 15

the answer is 1/3

Answer 16

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)

Answer 17

Thanks I got your point