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$\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}}=$
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1250 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{\sqrt{3}}$
We have,
$$
\begin{aligned}
& \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}} \\
& =\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}} \times \frac{\sqrt{a+2 x}+\sqrt{3 a}}{\sqrt{a+2 x}+\sqrt{3 a}} \\
& =\lim _{x \rightarrow a} \frac{(a+2 x)-3 a}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{a+2 x}+\sqrt{3 a}} \\
& =\lim _{x \rightarrow a} \frac{2 x-2 a}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{a+2 x}+\sqrt{3 a}} \\
& =2 \lim _{x \rightarrow a} \frac{(x-a)}{(x-a)} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{a+2 x}+\sqrt{3 a}} \\
& =2\left(\frac{\sqrt{a}+\sqrt{a}}{\sqrt{a+2 a}+\sqrt{3 a}}\right)=2 \frac{2 \sqrt{a}}{2 \sqrt{3} a}=\frac{2}{\sqrt{3}}
\end{aligned}
$$
$$
\begin{aligned}
& \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}} \\
& =\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}} \times \frac{\sqrt{a+2 x}+\sqrt{3 a}}{\sqrt{a+2 x}+\sqrt{3 a}} \\
& =\lim _{x \rightarrow a} \frac{(a+2 x)-3 a}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{a+2 x}+\sqrt{3 a}} \\
& =\lim _{x \rightarrow a} \frac{2 x-2 a}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{a+2 x}+\sqrt{3 a}} \\
& =2 \lim _{x \rightarrow a} \frac{(x-a)}{(x-a)} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{a+2 x}+\sqrt{3 a}} \\
& =2\left(\frac{\sqrt{a}+\sqrt{a}}{\sqrt{a+2 a}+\sqrt{3 a}}\right)=2 \frac{2 \sqrt{a}}{2 \sqrt{3} a}=\frac{2}{\sqrt{3}}
\end{aligned}
$$
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