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

Question
Limits and continuity
Evaluate the following limits
$$\lim_{x\rightarrow0}\frac{\sin7x}{\sin3x}$$

2020-11-11
We have to evaluate the limit:
$$\lim_{x\rightarrow0}\frac{\sin7x}{\sin3x}$$
We know that
$$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$$
if $$\lim_{x\rightarrow0}\frac{\sin ax}{x}$$
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.
$$\lim_{x\rightarrow0}\frac{\sin ax}{x}\times=\lim_{x\rightarrow0}\frac{\sin ax}{ax}\times a$$
$$=1\times a$$
$$=a$$
if $$\lim_{x\rightarrow0}\frac{\sin ax}{\sin bx}$$
then same case will be for numerator and denominator.
hence,
$$\lim_{x\rightarrow0}\frac{\sin ax}{\sin bx}=\frac{a}{b}$$
Finding given limit:
here, a=7
b=3
Hence,
$$\lim_{x\rightarrow0}\frac{\sin 7x}{\sin 3x}=\frac{7}{3}$$
Second method:
Multiplying and dividing by 7x for numerator and by 3x for denominator, we get
$$\lim_{x\rightarrow0}\frac{\sin 7x}{\sin 3x}\times\frac{7x}{7x}\times\frac{3x}{3x}=\lim_{x\rightarrow0}\frac{\frac{\sin7x}{7x}}{\frac{3x}{3x}}\times\frac{7x}{1}\times\frac{1}{3x}$$
$$=\frac{1}{1}\times\frac{7}{1}\times\frac{1}{3}$$
$$=\frac{7}{3}$$
Hence, value of the given limit is $$\frac{7}{3}$$

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