Q69 / 180 · COMEDK 2023 (Memory Based) · Mathematics
The value of $\lim _{x \rightarrow \infty}\left(\frac{x^2-2 x+1}{x^2-4 x+2}\right)^{2 x}$ is
- A$e^2$
- Correct answerB$e^4$
- C$e$
- D$e^{16}$
Solution
$\begin{aligned} & \text { Let } L=\lim _{x \rightarrow \infty}\left(\frac{x^2-2 x+1}{x^2-4 x+2}\right)^{2 x} \\ & \Rightarrow \ln L=\lim _{x \rightarrow \infty}\left[2 x \ln \left(\frac{x^2-2 x+1}{x^2-4 x+2}\right)\right] \\ & \Rightarrow \ln L=\lim _{x \rightarrow \infty}\left[2 x \ln \left(1+\frac{2 x-1}{x^2-4 x+2}\right)\right] \\ & \Rightarrow \ln L=\lim _{x \rightarrow \infty}\left[2 x\left(\frac{2 x-1}{x^2-4 x+2}-\frac{\left(\frac{2 x-1}{x^2-4 x+2}\right)^2}{2}+\right.\right. \\ & \left.+\frac{\left(\frac{2 x-1}{x^2-4 x+2}\right)^3}{3}\right) \ldots .\end{aligned}$
$\begin{aligned} & \Rightarrow \quad \ln L=\left(\lim _{x \rightarrow \infty} \frac{2 x(2 x-1)}{x^2-4 x+2}\right)-0+0 . \\ & \Rightarrow \quad \ln L=\lim _{x \rightarrow \infty} \frac{4 x^2\left(1-\frac{1}{2 x}\right)}{x^2\left(1-\frac{4}{x}+\frac{2}{x}\right)} \\ & \quad \ln L=\frac{4(1-0)}{(1-0+0)} \\ & \Rightarrow \quad \ln L=4 \\ & \Rightarrow \quad L=e^4\end{aligned}$
Hence, the correct answer is option: 'B'.