Limit (L'hopitals) trig functions

[whatlifeforme]

Homework Statement

use l'hopital's to evaluate the limit.

Homework Equations

limit (∅->0) ∅-3sin∅cos ∅
--------------------
tan∅- ∅

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

 

[SammyS]

Homework Statement

use l'hopital's to evaluate the limit.

Homework Equations

limit  (∅->0)        ∅-3sin∅cos ∅
                   --------------------
                        tan∅- ∅

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

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

 

[mfb]

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

 

[whatlifeforme]

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

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

 

[whatlifeforme]

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

using the product rule:

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

which means:

1-3(cos2x)
------------
(tanx)^2further: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.

 

[SammyS]

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.

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.