MCQ
$\mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {x + h} - \sqrt x }}{h} = $
- ✓$\frac{1}{{2\sqrt x }}$
- B$\frac{1}{{\sqrt x }}$
- C$2\sqrt x $
- D$\sqrt x $
✓
Answer
Correct option: A.
$\frac{1}{{2\sqrt x }}$
a
(a) $\mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sqrt {x + h} - \sqrt x }}{h} = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{{{(\sqrt {x + h} )}^2} - {{(\sqrt x )}^2}}}{{h\,(\sqrt {x + h} + \sqrt x )}} = \frac{1}{{2\sqrt x }}$.
(a) $\mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sqrt {x + h} - \sqrt x }}{h} = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{{{(\sqrt {x + h} )}^2} - {{(\sqrt x )}^2}}}{{h\,(\sqrt {x + h} + \sqrt x )}} = \frac{1}{{2\sqrt x }}$.
Aliter : Apply $L-$ Hospital rule,
$\mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sqrt {x + h} - \sqrt x }}{h} = \mathop {\lim }\limits_{h \to 0} \,\,\frac{1}{{2\sqrt {x + h} }} = \frac{1}{{2\sqrt x }}$.
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