If lim _{x} {rightarrow infty} frac{(sqrt{2 x | vedclass

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(D) Let $L = \lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8(P x^4-16)}{(x+\sqrt{x^2-2})^8+(x-\sqrt

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$\frac{1}{16}$

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(D) Let $L = \lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8(P x^4-16)}{(x+\sqrt{x^2-2})^8+(x-\sqrt{x^2-2})^8} = 1$. Divide the numerator and denominator by $x^4$ inside the terms or analyze the growth rates. As $x \rightarrow \infty$,$\sqrt{2x+1} + \sqrt{2x-1} \approx 2\sqrt{2x}$ and $\sqrt{2x+1} - \sqrt{2x-1} = \frac{2}{\sqrt{2x+1} + \sqrt{2x-1}} \approx \frac{2}{2\sqrt{2x}} = \frac{1}{\sqrt{2x}}$. Thus,$(\sqrt{2x+1} + \sqrt{2x-1})^8 \approx (2\sqrt{2x})^8 = 2^4 \cdot (2x)^4 = 16 \cdot 16x^4 = 256x^4$. Also,$(x+\sqrt{x^2-2})^8 \approx (2x)^8 = 256x^8$. The expression becomes $\lim _{x \rightarrow \infty} \frac{256x^4 + (\frac{1}{\sqrt{2x}})^8(Px^4-16)}{256x^8} = 1$. Since $(\frac{1}{\sqrt{2x}})^8 = \frac{1}{16x^4}$,the expression is $\lim _{x \rightarrow \infty} \frac{256x^4 + \frac{1}{16x^4}(Px^4-16)}{256x^8} = 1$. This simplifies to $\lim _{x \rightarrow \infty} \frac{256x^4 + \frac{P}{16} - \frac{1}{x^4}}{256x^8} = 1$. For the limit to be $1$,the leading coefficients must match. $256x^4 \cdot Px^4$ terms dominate,leading to $P = \frac{1}{16}$.

If $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8(P x^4-16)}{(x+\sqrt{x^2-2})^8+(x-\sqrt{x^2-2})^8} = 1$,then $P=$

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