finding the limit using l'Hospital's Rule if necessary?

Question

limit of (sinx - x)/ x^3 as x approaches 0

Answer 1

Sure.

Take the derivative of the numerator and denominator separately.

(cos(x) - 1) / 3x^2

Damn. It's still zero over zero as x approaches zero.

Are we allowed to apply L'Hopital's rule again?

-sin(x) / 6x

Still zero over zero.

Again?

-cos(x) / 6

The limit of this expression is -1/6 as x approaches zero.

Is that reasonable? Let's evaluate your original expression for x = 0.01. Yes, for x = .01 (radians), the expression evaluates to slightly greater than -1/6.

Let's evaluate it for -.01. Likewise, slightly greater than -1/6.

Answer 2

negative infinity...