trig limit problem

**kingsolomonsgrave** · October 21st 2012, 06:41 PM

for the problem above, is the idea that u can multiply as many times by 1 as you need to to get the form sinx/x or x/sinx?

**MarkFL** · October 21st 2012, 06:58 PM

Basically, yes. Until you learn L'Hôpital's rule, this is the way you must handle these types of problems, by algebraic manipulation.

**chiro** · October 21st 2012, 07:01 PM

The idea is that yes you multiply by 1, but remember that t is approaching zero which means that the limit has to be equal to 1 for these terms and in case it is.

If these new limits were not equal to 1 then you couldn't do the substitution.

**Prove It** · October 21st 2012, 07:03 PM

Yes that is exactly the idea. Though I would simply do this...

$\displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{3x}}{\sin{6x}} &= \lim_{x \to 0}\frac{\sin{3x}}{2\sin{3x}\cos{3x}} \\ &= \lim_{x \to 0}\frac{1}{2\cos{3x}} \\ &= \frac{1}{2 \cos {0}} \\ &= \frac{1}{2 \cdot 1} \\ &= \frac{1}{2} \end{align*}$

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