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) \[\dfrac{3}{{4\sqrt {2x} }}\]

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 .