# Thread: Limit of Trigonometric function

1. ## 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!

2. 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 $\frac{f}{g} \rightarrow \frac{f'(a)}{g'(a)}$.

3. 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:

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

For example, rewrite the first one as:

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{8x}} = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{4x}}$

4. Thanks, but I understand that I have to simplify them.

But for the second problem, I understand that

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

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

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

Is that correct?

5. Watch out, in the standard limit I gave, the angle has to be the same as the lineair factor. So rewrite:

$
\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

$
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
$

6. Originally Posted by TD!

$
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 $\frac{\sin 6x}{6}=1$?
You probably want to say, $\frac{\sin 6x}{6x}=1$ and you seem to forgot to state that.

7. Sorry but I"m getting a bit confused.

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

Sorry but I"m getting a bit confused.

Where does the 2 from the $2/[sin(2x)]$ and the 6 from $(tan6x)/6x$ come from?
He's multiplying and dividing by 2. Quick example:
$sin(2x)=\frac{2sin(2x)}{2}=2 \left ( \frac{sin(2x)}{2} \right )$.

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

9. I left out the x's to see if you were all paying attention