Trig Limit

**polymerase** · October 21st 2007, 06:06 AM

Hey

I know this question is rather simple but I just can't seem to get it...

$\displaystyle\lim_{x\rightarrow0}\frac{sin 5x}{sin 2x}-\frac{sin 3x}{4x}$

Thanks

**TD!** · October 21st 2007, 06:16 AM

You know the standard limit of sin(a)/a for a->0? It's 1. Rewrite:

$<br /> \lim_{x\rightarrow0}\left( \frac{\sin 5x}{\sin 2x}-\frac{\sin 3x}{4x} \right) =<br /> \lim_{x\rightarrow0}\left( \frac{5}{2} \frac{\sin 5x}{5x} \frac{2x}{\sin 2x}<br /> -\frac{3}{4}\frac{\sin 3x}{3x} \right)<br />$

Now take the limit of both terms and in each term, of all factors.

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