Question

What is $\mathop {\lim }\limits_{h \to 0}$ $\dfrac{{\sqrt {2x + 3h} - \sqrt {2x} }}{{2h}}$ equal to ?
(A) $\dfrac{1}{{2\sqrt {2x} }}$
(B) $\dfrac{3}{{\sqrt {2x} }}$
(C) $\dfrac{3}{{2\sqrt {2x} }}$
(D) $\dfrac{3}{{4\sqrt {2x} }}$

Answer 1

Hint : Here we can not put $h \to 0$ in denominator or in numerator as that will lead the limit to $\dfrac{0}{0}$ form, which is an indeterminate form . We can solve indeterminate forms using the factorization method or the rationalisation method . Here we have to use the rationalisation method to solve the limit .

Complete step-by-step answer :
Rationalisation method is particularly used when either the numerator or denominator or both involve expression consisting of square roots and substituting the value of $h$ the rational expression takes the form $\dfrac{0}{0}$ , $\dfrac{\infty }{\infty }$ .
In this method we have to rationalise the numerator as it only has square roots and after that simplify in such a way that $h$ gets canceled out both in numerator and denominator such a way that indeterminate form can be eliminated .
First we do the rationalisation part
\mathop {\lim }\limits_{h \to 0} $$$$\dfrac{{\sqrt {2x + 3h} - \sqrt {2x} }}{{2h}} (form $\dfrac{0}{0}$ )
We will multiply both numerator and denominator with the conjugate of the numerator which is $\sqrt {2x + 3h} + \sqrt {2x}$ to rationalise the numerator .
So the limit problem will become
\mathop {\lim }\limits_{h \to 0} $$$$\dfrac{{\left( {\sqrt {2x + 3h} - \sqrt {2x} } \right) \times \left( {\sqrt {2x + 3h} + \sqrt {2x} } \right)}}{{2h \times \left( {\sqrt {2x + 3h} + \sqrt {2x} } \right)}} (form $\dfrac{0}{0}$ )
After multiplying in numerator we get
= \mathop {\lim }\limits_{h \to 0} $$$$\dfrac{{2x + 3h - 2x}}{{2h \times \left( {\sqrt {2x + 3h} + \sqrt {2x} } \right)}} (form $\dfrac{0}{0}$ )
(As we use the algebraic formula $\left( {a + b} \right) \times \left( {a - b} \right) = {a^2} - {b^2}$;Here $a$ is $\sqrt {2x + 3h}$ and $b$is $\sqrt {2x}$ )
=$\mathop {\lim }\limits_{h \to 0}$ $\dfrac{{3h}}{{2h \times \left( {\sqrt {2x + 3h} + \sqrt {2x} } \right)}}$ (form $\dfrac{0}{0}$ )
= \mathop {\lim }\limits_{h \to 0} $$$$\dfrac{3}{{2 \times \left( {\sqrt {2x + 3h} + \sqrt {2x} } \right)}}
Now we can put $h \to 0$ in the numerator as it will not lead to indeterminate $\dfrac{0}{0}$ form .
So the sum becomes $\dfrac{3}{{2 \times \left( {\sqrt {2x + 0} + \sqrt {2x} } \right)}}$
= $\dfrac{3}{{4\sqrt {2x} }}$
So the correct option is (D).
So, the correct answer is “Option D”.

Note : While writing conjugate we need to be careful about where we put the opposite sign. We have to put the opposite sign between two square roots . We must remember the one and only purpose of all these calculations is to remove indeterminate form, so some time after rationalisation factorization may be needed to remove indeterminate form .