Question

# Use L'Hospital Rule to find the limits lim_{xrightarrow0}frac{sin mx}{sin nx}

Limits and continuity
Use L'Hospital Rule to find the limits
$$\lim_{x\rightarrow0}\frac{\sin mx}{\sin nx}$$

2020-10-28
We have to find the limits by using L 'Hospital's rule:
$$\lim_{x\rightarrow0}\frac{\sin mx}{\sin nx}$$
In L' Hospital's rule we differentiate numerator as well as denominator if they have the form $$\frac{0}{0}\ and\ \frac{\infty}{\infty}$$
After putting limits value we can say that this is $$\frac{0}{0}$$ form therefore we can successfully apply the rule so differentiating numerator and denominator with respect to x,
$$\lim_{x\rightarrow0}\frac{\sin mx}{\sin nx}=\lim_{x\rightarrow0}\frac{\frac{d(\sin mx)}{dx}}{\frac{d(\sin nx)}{dx}}$$
We know that
$$\frac{d(\sin ax)}{dx}=\cos ax\frac{d(ax)}{dx}$$
$$=\cos ax(a\frac {dx}{dx})$$
$$=a\cos ax$$
Solving further using above formula,
$$\lim_{x\rightarrow0}\frac{\frac{d(\sin mx)}{dx}}{\frac{d(\sin nx)}{dx}}=\lim_{x\rightarrow0}\frac{\cos mx(\frac{d(mx)}{dx})}{\cos nx(\frac{d(nx)}{dx})}$$
$$=\lim_{x\rightarrow0}\frac{\cos mx(m\frac{dx}{dx})}{\cos nx(n\frac{dx}{dx})}$$
$$=\lim_{x\rightarrow0}\frac{\cos mx(m\times1)}{\cos nx(n\times1)}$$
$$=\frac{m\cos m\times0}{n\cos n\times0}$$
$$=\frac{m\cos0}{n\cos0}$$
$$=\frac{m\times1}{n\times1}$$
$$=\frac{m}{n}$$
Hence, value of limit is $$=\frac{m}{n}$$