But What Is $b$ ?

$y=\large \displaystyle\lim_{a\to b} \left( \dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}} \right)$

Let $x=\sqrt{a}, z=\sqrt{b}$

The above limit can then be written as:

$y=\lim_{x\rightarrow z}\frac{x^{3}-z^{3}}{x-z}$

Using the formula of limits: $\lim_{x\rightarrow a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$

We get, $y=3z^{2}=\boxed{3b}$

$\large \displaystyle\lim_{a\to b} \left( \dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}} \right)$
Applying L'Hôpital's Rule , $\large \displaystyle\lim_{a\to b} \left(\dfrac{\frac{3}{2}\sqrt{a}}{\frac{1}{2\sqrt{a}}}\right)$
$\large \displaystyle \lim_{a \to b} \ ( 3a ) = \boxed{ 3b}$

Note : Here , b is a constant and its derivative with respect to a will be 0.

$\displaystyle \lim_{a\rightarrow b} \frac{a^{\frac{3}{2}}-b^{\frac{3}{2}}}{\sqrt{a}-\sqrt{b}}$

$\displaystyle \lim_{a\rightarrow b} \frac{ (\sqrt{a} -\sqrt{b} )( a+ \sqrt(ab) +b)} {\sqrt{a} - \sqrt{b}}$

$\displaystyle \lim_{a\rightarrow b} (a +\sqrt{ab} + b)$

$\displaystyle \Rightarrow b + \sqrt{b \times b} + b = \boxed{3b}$

$\begin{array}{l} \mathop {\lim }\limits_{a \to b} \frac{{a\sqrt a - b\sqrt b }}{{\sqrt a + \sqrt b }} = \\ = \mathop {\lim }\limits_{a \to b} \frac{{a\sqrt a - b\sqrt b }}{{\sqrt a + \sqrt b }} \times \frac{{\sqrt a - \sqrt b }}{{\sqrt a - \sqrt b }}\\ = \mathop {\lim }\limits_{a \to b} \frac{{{a^2} + a\sqrt {ab} - b\sqrt {ab} - {b^2}}}{{a - b}}\\ = \mathop {\lim }\limits_{a \to b} \frac{{{a^2} - {b^2} + \sqrt {ab} \left( {a - b} \right)}}{{a - b}}\\ = \mathop {\lim }\limits_{a \to b} \frac{{\left( {a - b} \right)\left( {a + b} \right) + \sqrt {ab} \left( {a - b} \right)}}{{a - b}}\\ = \mathop {\lim }\limits_{a \to b} \frac{{\left( {a - b} \right)\left( {a + b + \sqrt {ab} } \right)}}{{a - b}}\\ = \mathop {\lim }\limits_{a \to b} \left( {a + b + \sqrt {ab} } \right)\\ = b + b + \sqrt {b \times b} \\ = 3b \end{array}$

Remove square root and convert it into a^(3/2)-b(3^2)....from there we can see that the above expression is taking the form of formula for a cube- b cube expand it... Put the value and get the ans