Question

The value of the limit $\lim _{x \rightarrow 0}\left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right)^{2 m / x}$ is \((\mathrm{A})\) (B) 2 (C) \(\mathrm{e}^{6 \mathrm{~m}}\) (D) \(\ln 6 \mathrm{~m}\)

Short answer

Question: Determine the value of the limit: $\lim _{x \rightarrow 0}\left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right)^{2 m / x}$. Answer: (B) 2

Step-by-step solution

Definition of the exponential function

Recall that \(e^x = \lim_{n\to\infty} (1 + \frac{x}{n})^n\). Using this definition, we rewrite the limit expression: $$ \lim _{x \rightarrow 0}\left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right)^{2 m / x} = e^{\lim _{x \rightarrow 0} 2m \cdot \frac{\sin \frac{x}{m}+\cos \frac{3 x}{m} - 1}{x}} $$

Express using exponents

Introduce an auxiliary variable \(n = \frac{2m}{x}\) so that \(x = \frac{2m}{n}\) when \(n = \frac{2m}{x}\) and rewrite the limit: $$ \lim _{n \rightarrow \infty} 2m \cdot \frac{\sin \frac{2m}{nm}+\cos \frac{6 m}{n \cdot m} - 1}{\frac{2m}{n}} $$ Now substitute \(n\) back in place of the auxilary variable: $$ \lim _{n \rightarrow \infty} 2m \cdot \frac{\sin \frac{2m}{\frac{2m}{x}}+\cos \frac{6 m}{\frac{2m}{x} \cdot m} - 1}{\frac{2m}{\frac{2m}{x}}} $$

Apply L'Hôpital's Rule

Since the limit is in the indeterminate form of \(0/\infty\), we can apply L'Hôpital's Rule to find the limit. We obtain the derivatives of the numerator and denominator with respect to \(x\): $$ \frac{d}{dx}(\sin \frac{x}{m}+\cos \frac{3 x}{m} - 1) = \frac{1}{m} \cos \frac{x}{m} - \frac{3}{m} \sin \frac{3x}{m} $$ $$ \frac{d}{dx}(x) = 1 $$ Now we can take the limit as \(x \rightarrow 0\): $$ \lim _{x \rightarrow 0} \frac{\frac{1}{m} \cos \frac{x}{m} - \frac{3}{m} \sin \frac{3x}{m}}{1} = \frac{1}{m} \cos (0) - \frac{3}{m} \sin (0) = \frac{1}{m} $$ Therefore the limit expression becomes: $$ \lim _{x \rightarrow 0}\left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right)^{2 m / x} = e^{\lim _{x \rightarrow 0} 2m \cdot \frac{1}{m}} = e^2 $$ The answer is (B) 2.