Q95 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
The value of $\lim _\limits{x \rightarrow \infty}\left(\dfrac{x^2-2 x+1}{x^2-4 x+2}\right)^{2 x}$ is
- A$e^2$
- B$e^{16}$
- C$e$
- Correct answerD$e^4$
Solution
The limit is of the form $1^{\infty}$ as $x \rightarrow \infty$.
Let $L = \lim_{x \rightarrow \infty} \left(\dfrac{x^2-2x+1}{x^2-4x+2}\right)^{2x}$.
Using the formula $\lim_{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim_{x \rightarrow a} g(x)[f(x)-1]}$, we have:
$L = e^{\lim_{x \rightarrow \infty} 2x \left(\dfrac{x^2-2x+1}{x^2-4x+2} - 1\right)}$
$L = e^{\lim_{x \rightarrow \infty} 2x \left(\dfrac{x^2-2x+1 - (x^2-4x+2)}{x^2-4x+2}\right)}$
$L = e^{\lim_{x \rightarrow \infty} 2x \left(\dfrac{2x-1}{x^2-4x+2}\right)}$
$L = e^{\lim_{x \rightarrow \infty} \dfrac{4x^2-2x}{x^2-4x+2}}$
Dividing numerator and denominator by $x^2$:
$L = e^{\lim_{x \rightarrow \infty} \dfrac{4 - 2/x}{1 - 4/x + 2/x^2}} = e^{4/1} = e^4$.
Answer: $e^4$
Let $L = \lim_{x \rightarrow \infty} \left(\dfrac{x^2-2x+1}{x^2-4x+2}\right)^{2x}$.
Using the formula $\lim_{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim_{x \rightarrow a} g(x)[f(x)-1]}$, we have:
$L = e^{\lim_{x \rightarrow \infty} 2x \left(\dfrac{x^2-2x+1}{x^2-4x+2} - 1\right)}$
$L = e^{\lim_{x \rightarrow \infty} 2x \left(\dfrac{x^2-2x+1 - (x^2-4x+2)}{x^2-4x+2}\right)}$
$L = e^{\lim_{x \rightarrow \infty} 2x \left(\dfrac{2x-1}{x^2-4x+2}\right)}$
$L = e^{\lim_{x \rightarrow \infty} \dfrac{4x^2-2x}{x^2-4x+2}}$
Dividing numerator and denominator by $x^2$:
$L = e^{\lim_{x \rightarrow \infty} \dfrac{4 - 2/x}{1 - 4/x + 2/x^2}} = e^{4/1} = e^4$.
Answer: $e^4$
Hence, the correct answer is option: 'D'.