Answer
Using graphical, numerical and analytic methods, it can be seen that:
$\lim_{x\to0}\dfrac{\sqrt{x+2}-\sqrt{2}}{x}=\dfrac{\sqrt{2}}{4}\approx0.3536$
Work Step by Step
$\lim_{x\to0}\dfrac{\sqrt{x+2}-\sqrt{2}}{x}$
From the graph (shown in the answer section), $\lim_{x\to0}\dfrac{\sqrt{x+2}-\sqrt{2}}{x}$ can be estimated as $0.35$
To reinforce this estimation, elaborate a table (shown below), approaching $0$ from the left and from the right. It can be also seen that $\lim_{x\to0}\dfrac{\sqrt{x+2}-\sqrt{2}}{x}$ is $0.35$
Let's confirm these estimations by evaluating the limit by analytic methods:
$\lim_{x\to0}\dfrac{\sqrt{x+2}-\sqrt{2}}{x}=...$
Rationalize the numerator by multiplying the numerator and the denominator of this fraction by the conjugate of the numerator. Then, simplify:
$...=\lim_{x\to0}\dfrac{\sqrt{x+2}-\sqrt{2}}{x}\cdot\dfrac{\sqrt{x+2}+\sqrt{2}}{\sqrt{x+2}+\sqrt{2}}=...$
$...=\lim_{x\to0}\dfrac{(\sqrt{x+2})^{2}-(\sqrt{2})^{2}}{x(\sqrt{x+2}+\sqrt{2})}=...$
$...=\lim_{x\to0}\dfrac{x+2-2}{x(\sqrt{x+2}+\sqrt{2})}=\lim_{x\to0}\dfrac{x}{x(\sqrt{x+2}+\sqrt{2})}=...$
$...=\lim_{x\to0}\dfrac{1}{\sqrt{x+2}+\sqrt{2}}=...$
Evaluate applying direct substitution:
$...=\dfrac{1}{\sqrt{0+2}+\sqrt{2}}=\dfrac{1}{\sqrt{2}+\sqrt{2}}=\dfrac{1}{2\sqrt{2}}=...$
Rationalize the denominator to give the answer:
$...=\dfrac{1}{2\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{\sqrt{2}}{2(\sqrt{2})^{2}}=\dfrac{\sqrt{2}}{4}\approx0.3536$
