Q31 / 50 · Manipal (MET) 2022 · Mathematics
$\lim _{x \rightarrow 2} \frac{x^2+2^{2-x}-5}{2^{-x / 2}-2^{1-x}}$ is equal to _____
Correct answer: 12
Solution
We have,
$\lim _{x \rightarrow 2} \frac{2^x+2^{2-x}-5}{2^{-x / 2}-2^{1-x}} \quad\left[\frac{0}{0} \text { form }\right]$
So, apply L' Hospital's rule $=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}\left(2^x+2^{2-x}-5\right)}{\frac{d}{d x}\left(2^{-x / 2}-2^{1-x}\right)}$ $\begin{aligned} & =\lim _{x \rightarrow 2} \frac{2^x \ln 2-2^{2-x} \ln 2}{2^{1-x} \ln 2-\frac{2^{-\frac{x}{2}} \ln 2}{2}} \\ & =\lim _{x \rightarrow 2} \frac{2^{x+1}-2^{3-x}}{2^{2-x}-2^{-\frac{x}{2}}} \\ & =\frac{2^3-2^1}{2^0-2^{-1}}=\frac{8-2}{2-\frac{1}{2}}=\frac{12}{2-1}=12\end{aligned}$
So, apply L' Hospital's rule $=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}\left(2^x+2^{2-x}-5\right)}{\frac{d}{d x}\left(2^{-x / 2}-2^{1-x}\right)}$ $\begin{aligned} & =\lim _{x \rightarrow 2} \frac{2^x \ln 2-2^{2-x} \ln 2}{2^{1-x} \ln 2-\frac{2^{-\frac{x}{2}} \ln 2}{2}} \\ & =\lim _{x \rightarrow 2} \frac{2^{x+1}-2^{3-x}}{2^{2-x}-2^{-\frac{x}{2}}} \\ & =\frac{2^3-2^1}{2^0-2^{-1}}=\frac{8-2}{2-\frac{1}{2}}=\frac{12}{2-1}=12\end{aligned}$
Hence, the correct answer is 12.