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

Limits and continuity
asked 2020-10-27
Use L'Hospital Rule to find the limits
\(\lim_{x\rightarrow0}\frac{\sin mx}{\sin nx}\)

Answers (1)

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,
\(\lim_{x\rightarrow0}\frac{\frac{d(\sin mx)}{dx}}{\frac{d(\sin nx)}{dx}}=\lim_{x\rightarrow0}\frac{\cos mx(\frac{d(mx)}{dx})}{\cos nx(\frac{d(nx)}{dx})}\)
\(=\lim_{x\rightarrow0}\frac{\cos mx(m\frac{dx}{dx})}{\cos nx(n\frac{dx}{dx})}\)
\(=\lim_{x\rightarrow0}\frac{\cos mx(m\times1)}{\cos nx(n\times1)}\)
\(=\frac{m\cos m\times0}{n\cos n\times0}\)
Hence, value of limit is \(=\frac{m}{n}\)
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