Answer 1

\[\lim_{x \to 0} \left( \cos x + \sin x \right)^\frac{1}{x} \]
\[\text{ By adding and subtracting } 1, \text{ we get }: \]
\[ = \lim_{x \to 0} \left[ 1 + \cos x + \sin x - 1 \right]^\frac{1}{x} \]
\[\text{ Using the theorem given below }: \]
\[If \lim_{x \to a} f\left( x \right) = \lim_{x \to a} g\left( x \right) = 0 \text{ such that } \lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} \text{ exists, then } \lim_{x \to a} \left[ 1 + f\left( x \right) \right]^\frac{1}{g\left( x \right)} = e^\lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} . \]
\[Here: \]
\[ f\left( x \right) = \cos x + \sin x - 1\]
\[ g\left( x \right) = x\]
\[ \Rightarrow e^\lim_{x \to 0} \left( \frac{\cos x + \sin x - 1}{x} \right)\]
\[ \Rightarrow e^\lim_{x \to 0} \left[ \frac{\sin x}{x} - \frac{\left( 1 - \cos x \right)}{x} \right]\]
\[ \Rightarrow e^\lim_{x \to 0} \left( \frac{\sin x}{x} - \frac{2 \sin^2 \frac{x}{2}}{x} \right)\]
\[ \Rightarrow e^\lim_{x \to 0} \left( \frac{\sin x}{x} - \frac{2 \sin\left( \frac{x}{2} \right) \times \sin \left( \frac{x}{2} \right)}{2 \times \frac{x}{2}} \right)\]
\[ = e^{1 - 0} \]
\[ = e^1\]