Q109 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\lim _\limits{x \rightarrow 0} \dfrac{\sqrt{a+x}-\sqrt{a}}{x \sqrt{a(a+x)}}$ equals to
- A$\dfrac{1}{2}$
- B$a^{-\dfrac{3}{2}}$
- C$2 a^{-\dfrac{3}{2}}$
- Correct answerD$\dfrac{1}{2 a^{\dfrac{3}{2}}}$
Solution
Let the given limit be $L = \lim_{x \rightarrow 0} \dfrac{\sqrt{a+x} - \sqrt{a}}{x \sqrt{a(a+x)}}$.
Rationalizing the numerator by multiplying and dividing by $(\sqrt{a+x} + \sqrt{a})$:
$L = \lim_{x \rightarrow 0} \dfrac{(\sqrt{a+x} - \sqrt{a})(\sqrt{a+x} + \sqrt{a})}{x \sqrt{a(a+x)}(\sqrt{a+x} + \sqrt{a})}$
$L = \lim_{x \rightarrow 0} \dfrac{(a+x) - a}{x \sqrt{a(a+x)}(\sqrt{a+x} + \sqrt{a})}$
$L = \lim_{x \rightarrow 0} \dfrac{x}{x \sqrt{a(a+x)}(\sqrt{a+x} + \sqrt{a})}$
$L = \lim_{x \rightarrow 0} \dfrac{1}{\sqrt{a(a+x)}(\sqrt{a+x} + \sqrt{a})}$
Substituting $x = 0$ into the expression:
$L = \dfrac{1}{\sqrt{a(a+0)}(\sqrt{a+0} + \sqrt{a})}$
$L = \dfrac{1}{\sqrt{a^2}(\sqrt{a} + \sqrt{a})}$
$L = \dfrac{1}{a(2\sqrt{a})}$
$L = \dfrac{1}{2 a \cdot a^{\dfrac{1}{2}}} = \dfrac{1}{2 a^{\dfrac{3}{2}}}$
Answer: $\dfrac{1}{2 a^{\dfrac{3}{2}}}$
Rationalizing the numerator by multiplying and dividing by $(\sqrt{a+x} + \sqrt{a})$:
$L = \lim_{x \rightarrow 0} \dfrac{(\sqrt{a+x} - \sqrt{a})(\sqrt{a+x} + \sqrt{a})}{x \sqrt{a(a+x)}(\sqrt{a+x} + \sqrt{a})}$
$L = \lim_{x \rightarrow 0} \dfrac{(a+x) - a}{x \sqrt{a(a+x)}(\sqrt{a+x} + \sqrt{a})}$
$L = \lim_{x \rightarrow 0} \dfrac{x}{x \sqrt{a(a+x)}(\sqrt{a+x} + \sqrt{a})}$
$L = \lim_{x \rightarrow 0} \dfrac{1}{\sqrt{a(a+x)}(\sqrt{a+x} + \sqrt{a})}$
Substituting $x = 0$ into the expression:
$L = \dfrac{1}{\sqrt{a(a+0)}(\sqrt{a+0} + \sqrt{a})}$
$L = \dfrac{1}{\sqrt{a^2}(\sqrt{a} + \sqrt{a})}$
$L = \dfrac{1}{a(2\sqrt{a})}$
$L = \dfrac{1}{2 a \cdot a^{\dfrac{1}{2}}} = \dfrac{1}{2 a^{\dfrac{3}{2}}}$
Answer: $\dfrac{1}{2 a^{\dfrac{3}{2}}}$
Hence, the correct answer is option: 'D'.