# Limit of Trigonometric function

• Apr 1st 2006, 07:59 PM
Limit of Trigonometric function
I came into a few problems in which requires me to take limits of Trigonometric function. Now I have no idea how to start, but below is a couple of the problems.

1. Lim sin4x/8x
x->0

2. Lim tan6x/sin2x
x->0

I remember the rules for taking derivatives of Trigonometric functions, but I did not learn the foundation of these rules. So if anyone can help, please explain to me how to take limit of sin, cos, and tan.

Thank you very much!
• Apr 1st 2006, 09:28 PM
rgep
One useful rule in this content is L'Hopital's rule: if differentiable functions f and g both vanish at x=a, but the derivative g' does not, then $\displaystyle \frac{f}{g} \rightarrow \frac{f'(a)}{g'(a)}$.
• Apr 2nd 2006, 05:29 AM
TD!
Although L'Hopital works fine, these problems look like they have to be done without L'Hopital, by using a default limit you've probably seen, which is:

$\displaystyle \mathop {\lim }\limits_{\alpha \to 0} \frac{{\sin \alpha }}{\alpha } = 1$

For example, rewrite the first one as:

$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{8x}} = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{4x}}$
• Apr 2nd 2006, 09:15 AM
Thanks, but I understand that I have to simplify them.

But for the second problem, I understand that

$\displaystyle Lim (x->0) 1/cos x = 1$

therefore I simplify $\displaystyle (tan6x)/(sin2x)$ into $\displaystyle (sin6x)/[(cos6x)*(sin2x)]$

Take the limit (x->0) out of it, the answer is $\displaystyle 1/2$

Is that correct?
• Apr 2nd 2006, 09:27 AM
TD!
Watch out, in the standard limit I gave, the angle has to be the same as the lineair factor. So rewrite:

$\displaystyle \frac{{\tan \left( {6x} \right)}}{{\sin \left( {2x} \right)}} = \frac{{\sin \left( {6x} \right)}}{{\sin \left( {2x} \right)\cos \left( {6x} \right)}} = \frac{6}{2}\left( {\frac{{\sin \left( {6x} \right)}}{6}\frac{2}{{\sin \left( {2x} \right)}}\frac{1}{{\cos \left( {2x} \right)}}} \right)$

And then take the limit

$\displaystyle 3\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin \left( {6x} \right)}}{6}\frac{2}{{\sin \left( {2x} \right)}}\frac{1}{{\cos \left( {2x} \right)}}} \right) = 3 \cdot 1 \cdot 1 \cdot 1 = 3$
• Apr 2nd 2006, 09:55 AM
ThePerfectHacker
Quote:

Originally Posted by TD!

$\displaystyle 3\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin \left( {6x} \right)}}{6}\frac{2}{{\sin \left( {2x} \right)}}\frac{1}{{\cos \left( {2x} \right)}}} \right) = 3 \cdot 1 \cdot 1 \cdot 1 = 3$

WHAT?!?
The limit of $\displaystyle \frac{\sin 6x}{6}=1$?
You probably want to say, $\displaystyle \frac{\sin 6x}{6x}=1$ and you seem to forgot to state that.
• Apr 2nd 2006, 09:57 AM
Sorry but I"m getting a bit confused.

Where does the 2 from the $\displaystyle 2/[sin(2x)]$ and the 6 from $\displaystyle (tan6x)/6x$ come from?
• Apr 2nd 2006, 04:44 PM
topsquark
Quote:

Sorry but I"m getting a bit confused.

Where does the 2 from the $\displaystyle 2/[sin(2x)]$ and the 6 from $\displaystyle (tan6x)/6x$ come from?

He's multiplying and dividing by 2. Quick example:
$\displaystyle sin(2x)=\frac{2sin(2x)}{2}=2 \left ( \frac{sin(2x)}{2} \right )$.

Note the extra factors 6/2 on the outside of TD's expression.

(And, of course, ThePerfectHacker is correct, he really needed to be putting the x's in as well.)

-Dan
• Apr 3rd 2006, 12:41 AM
TD!
I left out the x's to see if you were all paying attention :D

:o