Find the limit for the given function.

**superduper1** · July 27th 2010, 04:28 PM

lim as $x$ approaches $0$ $\frac{\sin(5x)}{\sin(6x)}$

**Archie Meade** · July 27th 2010, 04:47 PM

Originally Posted by superduper1

lim as $x$ approaches $0$ $\frac{\sin(5x)}{\sin(6x)}$

Utilise the following limit

$\displaystyle\lim_{x\rightarrow0}\frac{sinx}{x}=1$

$\displaystyle\frac{sin5x}{sin6x}=\frac{5xsin5x}{5x }\ \frac{6x}{6xsin6x}=\frac{5x}{6x}\ \frac{sin5x}{5x}\ \frac{6x}{sin6x}$

**superduper1** · July 27th 2010, 05:24 PM

answer 5/6

**Archie Meade** · July 28th 2010, 04:13 AM

Yes,

$\displaystyle\lim_{6x\rightarrow\ 0}\frac{sin(6x)}{6x}=1$

and

$\displaystyle\lim_{5x\rightarrow\ 0}\frac{sin(5x)}{5x}=1$

The limit is often evaluated using the $\frac{sinx}{x}$ fraction,
though it's unnecessary if we write

$\displaystyle\lim_{y\rightarrow\ 0}siny=y$

Hence, $\displaystyle\lim_{x\rightarrow\ 0}\frac{sin(5x)}{sin(6x)}=\frac{\displaystyle\frac {5x}{5x}\lim_{5x\rightarrow\ 0}sin5x}{\displaystyle\frac{6x}{6x}\lim_{6x\righta rrow\ 0}sin6x}=\frac{5}{6}$

**Prove It** · July 28th 2010, 04:17 AM

Originally Posted by superduper1

lim as $x$ approaches $0$ $\frac{\sin(5x)}{\sin(6x)}$

Since this tends to $\frac{0}{0}$ you can use L'Hospital's Rule...

$\lim_{x \to 0}\frac{\sin{5x}}{\sin{6x}} = \lim_{x \to 0}\frac{\frac{d}{dx}(\sin{5x})}{\frac{d}{dx}(\sin{ 6x})}$

$= \lim_{x \to 0}\frac{5\cos{5x}}{6\cos{6x}}$

$= \frac{5\cos{(5\cdot 0)}}{6\cos{(6\cdot 0)}}$

$= \frac{5 \cdot 1}{6 \cdot 1}$

$= \frac{5}{6}$ .

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