Q101 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics
If $\lim _{x \rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8\left(P x^4-16\right)}{\left(x+\sqrt{x^2-2}\right)^8+\left(x-\sqrt{x^2-2}\right)^8}$ $=1$, then $P=$
- A16
- B64
- C$\frac{1}{64}$
- Correct answerD$\frac{1}{16}$
Solution
$\lim _{x \rightarrow \infty} \frac{\left[(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8\right]\left(P x^4-16\right)}{\left(x+\sqrt{x^2-2}\right)^8+\left(x-\sqrt{x^2-2}\right)^8}$
Dividing numerator and denominator by $x^8$ $\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\left[\left(\sqrt{2+\frac{1}{x}}+\sqrt{2-\frac{1}{x}}\right)^8+\left(\sqrt{2+\frac{1}{x}}-\sqrt{2-\frac{1}{x}}\right)^8\right]\left(P-\frac{16}{x^4}\right)}{\left(1+\sqrt{1-\frac{2}{x^2}}\right)^8+\left(1-\sqrt{1-\frac{2}{x^2}}\right)^8} \\ & =1 \end{aligned}$
Apply the limit $\begin{aligned} & \frac{\left[(\sqrt{2+0}+\sqrt{2-0})^8+(\sqrt{2}-\sqrt{2})^8\right] \mathrm{P}}{(1+1)^8+(1-1)^8}=1 \\ & \Rightarrow 2^4 \mathrm{P}=1 \Rightarrow \mathrm{P}=\frac{1}{16} \end{aligned}$
Dividing numerator and denominator by $x^8$ $\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\left[\left(\sqrt{2+\frac{1}{x}}+\sqrt{2-\frac{1}{x}}\right)^8+\left(\sqrt{2+\frac{1}{x}}-\sqrt{2-\frac{1}{x}}\right)^8\right]\left(P-\frac{16}{x^4}\right)}{\left(1+\sqrt{1-\frac{2}{x^2}}\right)^8+\left(1-\sqrt{1-\frac{2}{x^2}}\right)^8} \\ & =1 \end{aligned}$
Apply the limit $\begin{aligned} & \frac{\left[(\sqrt{2+0}+\sqrt{2-0})^8+(\sqrt{2}-\sqrt{2})^8\right] \mathrm{P}}{(1+1)^8+(1-1)^8}=1 \\ & \Rightarrow 2^4 \mathrm{P}=1 \Rightarrow \mathrm{P}=\frac{1}{16} \end{aligned}$
Hence, the correct answer is option: 'D'.