# Limit of Trigonometric f'n as x approaches 0

• joshmdmd
In summary, the limit as x tends to zero of (sinx-tanx)/x^3 can be solved by using the Taylor series of sin and tan and taking the limit as x approaches zero. Alternatively, one can multiply the top and bottom of the fraction by (\sin(x)+\tan(x)) and simplify to obtain the solution.

## 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.

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.

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.

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 $(\sin(x)+\tan(x))$.