How do you evaluate the limit sin(ax)/sin(bx) as x approaches 0?

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Explanation:

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Eddie Share
Sep 2, 2016

$= \frac{a}{b}$

Explanation:

${\lim}_{x \to 0} \sin \frac{a x}{\sin} \left(b x\right)$ is in $\frac{0}{0}$ indeterminate form so we can use l'Hopital's rule

$= {\lim}_{x \to 0} \frac{a \cos \left(a x\right)}{b \cos \left(b x\right)}$

we can lift out the contant term and note that the limit of the quotient is the quotient of the limits where the limits are known

$= \frac{a}{b} \frac{{\lim}_{x \to 0} \cos \left(a x\right)}{{\lim}_{x \to 0} \cos \left(b x\right)}$

$= \frac{a}{b} {\lim}_{x \to 0} \frac{1}{1} = \frac{a}{b}$

Another approach is to head for known limit ${\lim}_{z \to 0} \frac{\sin z}{z} = 1$ and to proceed as follows

${\lim}_{x \to 0} \sin \frac{a x}{\sin} \left(b x\right)$

$= {\lim}_{x \to 0} \sin \frac{a x}{x} \frac{x}{\sin \left(b x\right)}$

$= {\lim}_{x \to 0} \frac{\sin \frac{a x}{x}}{\frac{\sin \left(b x\right)}{x}}$

$= \frac{a}{b} {\lim}_{x \to 0} \frac{\sin \frac{a x}{a x}}{\frac{\sin \left(b x\right)}{b x}}$

$= \frac{a}{b} \frac{{\lim}_{x \to 0} \sin \frac{a x}{a x}}{{\lim}_{x \to 0} \frac{\sin \left(b x\right)}{b x}}$

with subs $u = a x , v = w z$

$= \frac{a}{b} \frac{{\lim}_{u \to 0} \sin \frac{u}{u}}{{\lim}_{v \to 0} \frac{\sin \left(v\right)}{v}}$

$= \frac{a}{b} \cdot \frac{1}{1}$

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