Q246 / 270 · NDA 2018 (Phase 1) · Mathematics

What is $\lim _{h \rightarrow 0} \frac{\sqrt{2 x+3 h}-\sqrt{2 x}}{2 h}$ equal to?

  • A
    $\frac{1}{2 \sqrt{2 x}}$
  • B
    $\frac{3}{\sqrt{2 \mathrm{x}}}$
  • C
    $\frac{3}{2 \sqrt{2 x}}$
  • Correct answer
    D
    $\frac{3}{4 \sqrt{2 x}}$

Solution

Let $L=\lim _{h \rightarrow 0} \frac{\sqrt{2 x+3 h}-\sqrt{2 x}}{2 h}$ Rationalise the numerator, we get: $L=\lim _{h \rightarrow 0}\left(\frac{\sqrt{2 x+3 h}-\sqrt{2 x}}{2 h} \times \frac{\sqrt{2 x+3 h}+\sqrt{2 x}}{\sqrt{2 x+3 h}+\sqrt{2 x}}\right)$ $\begin{aligned} & =\lim _{h \rightarrow 0} \frac{2 x+3 h-2 x}{2 h(\sqrt{2 x+3 h}+\sqrt{2 x})} \\ & =\frac{3}{2(\sqrt{2 x+0}+\sqrt{2 x})}=\frac{3}{4 \sqrt{2 x}}\end{aligned}$

Hence, the correct answer is option: 'D'.

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