lim (sin(x) - x) / x^3
so i tried splitting it up into sin(x) / x^3 - x / x^3. sin(x) / x^3 = (1/x^2)(sin(x) / x) and the limit as x approaches 0 of sin(x)/x = 1 so the first term just becomes 1/x^2. for the second term i canceled out an x from the top and bottom so i also got 1/x^2. so 1/x^2 - 1/x^2 = 0, so i got that the limit is equal to 0. but when i used L'Hospital's Rule on it, i got a completely different answer, -1/6. was the algebraic way i was doing it wrong? please help.
Your algebra is off . . .
Presently, we have: . . . . an indeterminate form.
We can write this as: . . . . an indeterminate form -- again!
We can apply L'Hopital: . . . . indeterminate
At this point, we can: .
. . L'Hopital again: . . . . indeterminate
. . L'Hopital again: .
. . We have: .
. . Multiply by .
. . Take the limit: .