Q.Evaluate : $\lim_{x \to 0} \dfrac{\sqrt{2+x} - \sqrt{2-x}}{2x}$.

Rationalise the numerator; the numerator becomes $2x$ which cancels the $2x$ in the denominator, giving $\dfrac{1}{\sqrt{2+x}+\sqrt{2-x}}\to\dfrac{1}{2\sqrt2}=\dfrac{\sqrt2}{4}$.

A $\frac{0}{0}$ limit from the Differential Calculus unit of the Tamil Nadu HSC Class-11 Business Mathematics syllabus.

Step 1 — Rationalise. Multiply numerator and denominator by the conjugate $\sqrt{2+x}+\sqrt{2-x}$:

$\frac{\sqrt{2+x}-\sqrt{2-x}}{2x}\cdot\frac{\sqrt{2+x}+\sqrt{2-x}}{\sqrt{2+x}+\sqrt{2-x}}=\frac{(2+x)-(2-x)}{2x\left(\sqrt{2+x}+\sqrt{2-x}\right)}.$

Step 2 — Simplify the numerator. $(2+x)-(2-x)=2x$, so …