Question

Answer 1

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