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$\lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+b x+x}-a x\right)=$
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Verified Answer
The correct answer is:
$\frac{b+1}{2 a}$
We have, $\lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+b x+x}-a x\right)$
$$
\begin{aligned}
&=\lim _{x \rightarrow \infty}\left(\left(\sqrt{a^{2} x^{2}+b x+x}-a x\right) \times \frac{\left(\sqrt{a^{2} x^{2}+b x+x}+a x\right)}{\left(\sqrt{a^{2} x^{2}+b x}+x+a x\right)}\right) \\
&=\lim _{x \rightarrow \infty}\left(\frac{\left(a^{2} x^{2}+b x+x\right)-a^{2} x^{2}}{x\left\{\sqrt{a^{2}+\frac{b}{x}+\frac{1}{x}}+a\right\}}\right) \\
&=\lim _{x \rightarrow \infty}\left(\frac{(b+1) x}{x\left\{\sqrt{a^{2}+\frac{b}{x}+\frac{1}{x}}+a\right\}}\right)=\frac{b+1}{2 a}
\end{aligned}
$$
$$
\begin{aligned}
&=\lim _{x \rightarrow \infty}\left(\left(\sqrt{a^{2} x^{2}+b x+x}-a x\right) \times \frac{\left(\sqrt{a^{2} x^{2}+b x+x}+a x\right)}{\left(\sqrt{a^{2} x^{2}+b x}+x+a x\right)}\right) \\
&=\lim _{x \rightarrow \infty}\left(\frac{\left(a^{2} x^{2}+b x+x\right)-a^{2} x^{2}}{x\left\{\sqrt{a^{2}+\frac{b}{x}+\frac{1}{x}}+a\right\}}\right) \\
&=\lim _{x \rightarrow \infty}\left(\frac{(b+1) x}{x\left\{\sqrt{a^{2}+\frac{b}{x}+\frac{1}{x}}+a\right\}}\right)=\frac{b+1}{2 a}
\end{aligned}
$$
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