Question

Evaluate: $\underset{x\to 2}{\lim }\left(\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\right)$

• A. $\frac{1}{8\sqrt{3}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $8\sqrt{3}$
• D. $\sqrt{3}$

Answer 1

The correct option is A

$\frac{1}{8\sqrt{3}}$

Explanation for the correct option:

Find the value of given equation:

Given,$\underset{x\to 2}{\lim }\left(\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\right)$

Consider the given Equation as

$$\begin{gathered} I=\underset{x\to 2}{\lim }\left(\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\right) \\ I=\left(\frac{\sqrt{1+\sqrt{2+2}}-\sqrt{3}}{2-2}\right) \\ I=\left(\frac{\sqrt{1+2}-\sqrt{3}}{0}\right) \\ I=\left(\frac{\sqrt{3}-\sqrt{3}}{0}\right) \\ I=\frac{0}{0}\left[\text{form}\right] \end{gathered}$$

Using the L' Hospital rule

$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

$$\begin{gathered} I=\underset{x\to 2}{\lim }\left(\frac{{\displaystyle \frac{1}{2}}{\left(1+\left(\sqrt{2+x}\right)\right)}^{{\displaystyle \frac{-1}{2}}}\times {\displaystyle \frac{1}{2}}{\left(2+x\right)}^{{\displaystyle \frac{-1}{2}}}}{1}\right)\left[\text{Apply chain rule}\right] \\ I=\frac{1}{2}{\left(1+\left(\sqrt{2+2}\right)\right)}^{\frac{-1}{2}}\times \frac{1}{2}{\left(2+2\right)}^{\frac{-1}{2}} \\ I=\frac{1}{4}\times {\left(1+\left(\sqrt{4}\right)\right)}^{\frac{-1}{2}}\times {\left(4\right)}^{\frac{-1}{2}} \\ I=\frac{1}{4}\times {\left(3\right)}^{\frac{-1}{2}}\times \frac{1}{{\left(4\right)}^{\frac{1}{2}}} \\ I=\frac{1}{4}\times \frac{1}{\sqrt{3}}\times \frac{1}{2} \\ I=\frac{1}{8\sqrt{3}} \end{gathered}$$

Hence, option (A) is the correct answer.