# Evaluate the following limits lim_{xrightarrow0}frac{sin7x}{sin3x}

Evaluate the following limits
$\underset{x\to 0}{lim}\frac{\mathrm{sin}7x}{\mathrm{sin}3x}$
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casincal
We have to evaluate the limit:
$\underset{x\to 0}{lim}\frac{\mathrm{sin}7x}{\mathrm{sin}3x}$
We know that
$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$
if $\underset{x\to 0}{lim}\frac{\mathrm{sin}ax}{x}$
in this case we need to multiply and divide by a since we do make same in the denominator as in the angle of sine.
$\underset{x\to 0}{lim}\frac{\mathrm{sin}ax}{x}×=\underset{x\to 0}{lim}\frac{\mathrm{sin}ax}{ax}×a$
$=1×a$
$=a$
if $\underset{x\to 0}{lim}\frac{\mathrm{sin}ax}{\mathrm{sin}bx}$
then same case will be for numerator and denominator.
hence,
$\underset{x\to 0}{lim}\frac{\mathrm{sin}ax}{\mathrm{sin}bx}=\frac{a}{b}$
Finding given limit:
here, a=7
b=3
Hence,
$\underset{x\to 0}{lim}\frac{\mathrm{sin}7x}{\mathrm{sin}3x}=\frac{7}{3}$
Second method:
Multiplying and dividing by 7x for numerator and by 3x for denominator, we get
$\underset{x\to 0}{lim}\frac{\mathrm{sin}7x}{\mathrm{sin}3x}×\frac{7x}{7x}×\frac{3x}{3x}=\underset{x\to 0}{lim}\frac{\frac{\mathrm{sin}7x}{7x}}{\frac{3x}{3x}}×\frac{7x}{1}×\frac{1}{3x}$
$=\frac{1}{1}×\frac{7}{1}×\frac{1}{3}$
$=\frac{7}{3}$
Hence, value of the given limit is $\frac{7}{3}$
Jeffrey Jordon