Question

$\underset{x\to 0}{\lim }{\left(\cos x+a\sin bx\right)}^{1/x}$

Answer 1

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