Question

Find the limit of sinx/x^3 if x approaches to 0-.

Answer 1

This limit diverges to positive infinity. In this case, as x->0, sin(x) can be replaced by x. So we get
\[ \lim_{x->0^-}=\lim_{x->0^-}\frac{\sin x}{x^3}=\lim_{x->0^-}\frac{1}{x^2}=+\infty\]

Answer 2

How did you get 1/x^2?

Answer 3

As sin(x) approaches 0, the approximation sin(x)=x becomes better and better, so I simply got 1/x^2 by replacing sin(x) by x.

Answer 4

If that's too confusing Ideal, consider that this problem is a good candidate for L'Hopital's Rule :o

Answer 5

I did L'Hopital's Rule and got -1/6, but that's not the answer.

Answer 6

Indeed, in general when you have 0/0 as the "limit" you can use L'Hopital's Rule:
\[ \lim_{x\rightarrow0}\frac{\sin x}{x^3}=\lim_{x\rightarrow0}\frac{\cos x}{3x^2} \]
The numerator tends to 1, a finite value, the denominator tends to 0, which means that the limit blows up, i.e. diverges to infinity.

Answer 7

\[\Large L'H \to\qquad=\lim_{x\to0^-}\frac{\cos x}{3x^2}\]After you apply L'Hop once, you shouldn't be able to perform it again.
We're no longer getting an indeterminate form that allows us to perform it again.

Answer 8

It's not related to ode, my prof gave us this assignment to refresh our brains.

Answer 9

Thanks guys.

Answer 10

this would also be an excellent candidate for a power series representation
you would get the answer instantly

Answer 11

@Idealist here. I need place to answer, right?

Answer 12

do as they teach you: L'hopital rule. do you know that rule? just take derivative both numerator and denominator. if the result still form the form of 0/0 or infty/infty , take l'hopital rule again. ok?

Answer 13

Okay.

Answer 14

show me your work, stubborn girl. it's not hard.

Answer 15

It's in my new post.

Answer 16

where? just tag me

Answer 17

oh, I saw it, but too many people there. I don't show up