## anonymous 4 years ago 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.

1. anonymous

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

2. Zarkon

how about a power series solution :)

3. anonymous

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

4. Zarkon

lol

5. Zarkon

go directly to jail

6. anonymous

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

7. Zarkon

power series gives the quickest solution

8. anonymous

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

9. anonymous

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

12. 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

13. ash2326

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

14. ash2326

Zarkon what do you think , is my method correct?

15. Zarkon

16. 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)

17. ash2326