If the function given by $f(x)=\left(\frac{4 x+1}{1-4 x} ight)^{\frac{1}{x}}$ for $x
eq o$ is continuous at $x=0$, then the value of $\mathrm{f}(\mathrm{o})$ is

Question

If the function given by $f(x)=\left(\frac{4 x+1}{1-4 x}ight)^{\frac{1}{x}}$ for $x 
eq o$ is continuous at $x=0$, then the value of $\mathrm{f}(\mathrm{o})$ is

Aicharya · Accepted Answer

$\begin{aligned} \lim _{x ightarrow 0}\left(\frac{1+4 x}{1-4 x}ight)^{\frac{1}{x}} &=\frac{\lim _{x ightarrow 0}(1+4 x)^{\frac{1}{x}}}{\lim _{x ightarrow 0}(1-4 x)^{\frac{1}{x}}}=\frac{\left[\lim _{x ightarrow 0}(1+4 x)^{\frac{1}{4 x}}ight]^{4}}{\left[\lim _{x ightarrow 0}(1-4 x)^{\frac{-1}{4 x}}ight]^{-4}}=\frac{e^{4}}{e^{-4}} \ &=e^{8} \end{aligned}$

If the function given by $f(x)=\left(\frac{4 x+1}{1-4 x}\right)^{\frac{1}{x}}$ for $x \neq o$ is continuous…

Solution