Question

For $x >0,\underset{x \rightarrow 0} {\text{Lim}} \left((\tan x)^{\frac{1}{x}}+(1+\sin x)^x+(\cot x)^x\right)$ is equal to

• A. 1
• B. 0
• C. 2
• D. non existent

Answer 1

Answer: (C)

$ \underset{x \rightarrow 0} {\text{Lim}}(\tan x)^{\frac{1}{x}}=\underset{x \rightarrow 0} {\text{Lim}}(\tan h)^{\frac{1}{h}}=0$
$\underset{x \rightarrow 0} {\text{Lim}}(1+\sin x)^x=(1+0)^0=1$
Let $l=\underset{x \rightarrow 0} {\text{Lim}}(\cot x )^{ x }\left(\infty^0\right.$ form $)$
$\therefore \ln l=\underset{x \rightarrow 0} {\text{Lim}} x \ln (\cot x)=\underset{x \rightarrow 0} {\text{Lim}} \frac{\ln (\cot x)}{\frac{1}{x}}$
$=\underset{x \rightarrow 0} {\text{Lim}} \frac{\frac{-\operatorname{cosec}^2 x}{\cot x}}{\frac{-1}{x^2}}=
\underset{x \rightarrow 0} {\text{Lim}}\left(\frac{x^2}{\sin ^2 x}\right) \tan x=0$
(Using L'Hospital's Rule)
$\Rightarrow l=1$
Hence $\underset{x \rightarrow 0} {\text{Lim}}\left((\tan x)^{\frac{1}{x}}+(1+\sin x)^x+(\cot x)^x\right)=0+1+1=2$