Q31 / 50 · Manipal (MET) 2023 · Mathematics
The value of $f(x)=\lim _{x \rightarrow 0}\left(\frac{8^x+2^x+4^x}{3}\right)^{2 / x}$, then $f(x)^{\prime}$ equals to
Correct answer: 16
Solution
$y=\lim _{x \rightarrow 0}\left(\frac{8^x+2^x+4^x}{3}\right)^{2 / x}$
$\begin{aligned} & \Rightarrow \log y=\lim _{x \rightarrow 0} \frac{2}{x} \log \left(\frac{8^x+2^x+4^x}{3}\right) \\ & =2 \lim _{x \rightarrow 0} \frac{1}{x}\left[\log \left(8^x+2^x+4^x\right)-\log 3\right]\end{aligned}$
$=2 \lim _{x \rightarrow 0} \frac{\left[\log \left(8^x+2^x+4^x\right)-\log 3\right]}{x} \quad\left[\frac{0}{0}\right.$ form $]$
On applying L' Hospital rule, we get
$\begin{aligned} & =2 \lim _{x \rightarrow 0} \frac{\left[8^x \log 8+2^x \log 2+4^x \log 4\right]}{8^x+2^x+4^x} \\ & =2\left[\frac{\log 8+\log 2+\log 4}{3}\right]\end{aligned}$
$=\frac{2}{3}[3 \log 2+\log 2+2 \log 2] \quad\left[\log (m)^n=n \log m\right]$
$\begin{aligned} & =\frac{2}{3} \times 6 \log 2 \\ & \log y=4 \log 2=\log 16 \Rightarrow y=16\end{aligned}$
Hence, the correct answer is 16.