Limit of Trigonometric f'n as x approaches 0

[joshmdmd]

Homework Statement

Find the following limit as x tends to zero
(idk how to code this stuff properly so the lim is lim as x->0)
lim (sinx-tanx)/x^3

Homework Equations

lim(sinx)/x = 1
lim(cosx) = 1
lim(1-cosx)/x = 0

The Attempt at a Solution

attempted to multiply by conjugate
attempted to remove the root from the top and the bottom
no idea how to do this.

 

[Coto]

I'm not sure what you are allowed to do here, but one way to solve this limit would be to consider the Taylor series of sin and tan about the point x = 0.

In particular, $\sin(x) \approx x - x^3/3! + x^5/5! - ...$ and $\tan(x) \approx x + x^3/3 + 2x^5/5 + ...$. Hence, subtracting these two series, gives:

$$\sin(x) - \tan(x) \approx -(x^3/3! + x^3/3) + O(x^5)$$

Substituting this into your function (i.e. dividing the above by $x^3$) and taking the limit as that goes to zero will give you an answer.

 

[joshmdmd]

nope. it has to be done without L'Hospital's rule, and we learn the taylor / maclaurin series next semester?. My attempts to rearrange it only get down to x on the bottom, can't seem to shake it out.

 

[Coto]

Okay... here is a slightly more cumbersome way,

Notice,
$$\frac{\sin(x) - \tan(x)}{x^3} = -\left(\frac{\sin(x)}{x}\right)^3\frac{1}{\cos^2(x) + \cos(x)}$$.

A simple application of limit laws will give you the solution from that stage.

Hint:
To obtain the above equality, start by multiplying the top and bottom of your fraction by $(\sin(x)+\tan(x))$.