calculating limits invloving trig functions

**juventinoalex** · September 1st 2010, 11:55 AM

Here is the problem.

lim as x approaches 0 of (tan4x/sin2x)

So far I have done this.

(sin7x/cos7x)/(sin2x)

(sin7x/cos7x) multiplied by (1/sin2x)

I'm not sure where to go from here. Suggestions? Thanks.

juventinoalex

**bondesan** · September 1st 2010, 12:01 PM

Try to use the fundamental trigonometric limit $\lim_{x\to 0}\dfrac{sin(x)}{x} = 1$ and its consequences when it comes to calculate the limit for the tangent.

**juventinoalex** · September 1st 2010, 12:12 PM

I still don't think I understand. Could someone explain further?

Thanks,
juventinoalex

**Defunkt** · September 1st 2010, 12:36 PM

1) Is it tan(4x) or tan(7x)?

2) $\frac{sin(7x)}{sin(2x)cos(7x)} = sin(7x) \cdot \frac{7x}{7x} \cdot \frac{1}{sin(2x)} \cdot \frac{2x}{2x} \cdot \frac{1}{cos(7x)}$ $= \frac{7}{2} \cdot \frac{sin(7x)}{7x} \cdot \frac {2x}{sin(2x)} \cdot \frac{1}{cos(7x)}$

**Plato** · September 1st 2010, 12:40 PM

Originally Posted by juventinoalex

I still don't think I understand. Could someone explain further?

$\displaystyle \tan(4x)=\frac{\sin(4x)}{\cos(4x)}=\frac{2\sin(2x) \cos(2x)}{\cos(4x)}$ .

So $\displaystyle \frac{\tan(4x)}{\sin(2x)}=\frac{2\cos(2x)}{\cos(4x )}}$

**juventinoalex** · September 1st 2010, 12:41 PM

Sorry for the typo. It is tan4x

**Plato** · September 1st 2010, 12:45 PM

See my reply #5.

**juventinoalex** · September 1st 2010, 01:06 PM

Got it now. Thanks so much.

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