Find the value of \(\lim _{x \rightarrow 0} \frac{\sin \left(x^m\right)}{(\sin x)^n}\), given that \(n < m\)
Find the value of \(\lim _{x \rightarrow 0} \frac{\sin \left(x^m\right)}{(\sin x)^n}\), given that \(n < m\)
- 2
- 1
- 0
- \(\infty\)
Solution
$\begin{aligned}
\lim_{x \rightarrow 0} & \frac{\sin \left(x^m\right)}{(\sin x)^n} ;(n < m) \\
& =\lim_{x \rightarrow 0} \frac{\frac{\sin x^m}{x^m} x^m}{\left(\frac{\sin x}{x} x\right)^n}=\lim_{x \rightarrow 0} \frac{\frac{\sin x^m}{x^m}}{\left(\frac{\sin x}{x}\right)^n} x^{m-n} \\
& =\frac{1}{1} \times 0 \quad \{\because m > n\} \\
& =0
\end{aligned}$
Hence, option (c) is correct.
Asked in: AP EAMCET 2020 (21 Sep Shift 2)
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