# Limit (L'hopitals) trig functions

1. Mar 3, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
use l'hopital's to evaluate the limit.

2. Relevant equations
limit (∅->0) ∅-3sin∅cos ∅
--------------------
tan∅- ∅

3. The attempt at a solution
i take the derivatives of the top and bottom, and use trig identity for sec^2x - 1 to tan^x.
i get 6(cosx)^2 annd get the value of 6. but the answer is -∞.

6(sinx)^2
----------
(tanx)^2

6(cosx)^2

2. Mar 3, 2013

### SammyS

Staff Emeritus
What is the derivative of -3sin(θ)cos(θ) ? (It's not 6 sin2(θ) . )

3. Mar 3, 2013

### Staff: Mentor

The same fraction, but easier to read with LaTeX:
$$\lim_{x \to 0} \frac{x-3\sin(x)\cos(x)}{\tan(x)-x}$$

4. Mar 4, 2013

### whatlifeforme

those values were actually taking out of a solution manual, but i can check again.

5. Mar 4, 2013

### whatlifeforme

using the product rule:

-3( (cosx)^2 - (sinx)^2 ) == cos2x

which means:

1-3(cos2x)
------------
(tanx)^2

further:

1-(3/2)sin(2x)
-----------------
2tanx * (secx)^2

i'm still not getting to -infinity, and it seems to be getting complicated in the denominator.

Last edited: Mar 4, 2013
6. Mar 4, 2013

### SammyS

Staff Emeritus
Yes, using the product rule, $\displaystyle \ \ \frac{d}{dx}\left(\sin(xcos(x)\right)=\cos^2(x)-\sin^2(x)\ \$ and that's equal to $\ \cos(2x)\ .\$

So, you have $\displaystyle \ \lim_{x \to 0} \frac{x-3\sin(x)\cos(x)}{\tan(x)-x}= \lim_{x \to 0}\,\frac{1-3\cos(2x)}{\tan^2(x)}\,,\$ which is not of indeterminate form, and gives the result you're looking for.