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प्रश्न
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
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उत्तर
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^\frac{1}{x} \]
\[ = \lim_{x \to 0} \left[ 1 + \cos x + a \sin bx - 1 \right]^\frac{1}{x} \]
\[\text{ Using the theoremgiven below }:\]
\[If \lim_{x \to a} f\left( x \right) = \lim_{x \to a} g\left( x \right) = 0 \text{ such that } \lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} \text{ exists, then } \lim_{x \to a} \left[ 1 + f\left( x \right) \right]^\frac{1}{g\left( x \right)} = e^\lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} . \]
\[Here: \]
\[ f\left( x \right) = \cos x + a \sin bx - 1\]
\[ g\left( x \right) = x\]
\[ \Rightarrow e^\lim_{x \to 0} \left[ \frac{\cos x + a \sin bx - 1}{x} \right] \]
\[ = e^\lim_{x \to 0} \left[ \frac{b \times a \sin bx}{bx} - \frac{\left( 1 - \cos x \right)}{x} \right] \]
\[ = e^\lim_{x \to 0} \left( \frac{ab \sin bx}{bx} - \frac{2 \sin^2 \frac{x}{2}}{x} \right) \]
\[ = e^{ab}\]
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