Find the limit:
[tex]\[
\lim _{x \rightarrow p} \frac{x \sqrt{x} - p \sqrt{p}}{\sqrt{x} - \sqrt{p}} = \cdots
\][/tex]

(A) [tex]\( p \sqrt{p^{-}} \)[/tex]
(B) [tex]\( 3p \)[/tex]
(C) [tex]\( p \)[/tex]
(D) [tex]\( 3 \sqrt{p} \)[/tex]
(E) [tex]\( 2p \)[/tex]



Answer :

To evaluate the limit
[tex]\[ \lim _{x \rightarrow p} \frac{x \sqrt{x}-p \sqrt{p}}{\sqrt{x}-\sqrt{p}}, \][/tex]
we start by observing the numerator and the denominator separately and how they behave as [tex]\( x \)[/tex] approaches [tex]\( p \)[/tex].

Notice that when [tex]\( x \)[/tex] approaches [tex]\( p \)[/tex], both the numerator [tex]\( x \sqrt{x} - p \sqrt{p} \)[/tex] and the denominator [tex]\( \sqrt{x} - \sqrt{p} \)[/tex] approach zero. This suggests that the expression is in an indeterminate form [tex]\( \frac{0}{0} \)[/tex], which typically calls for techniques such as algebraic manipulation or L'Hopital's Rule to resolve.

First, let’s rewrite the expression more conveniently. Define
[tex]\[ f(x) = x \sqrt{x} = x^{3/2}, \][/tex]
so the limit becomes
[tex]\[ \lim_{x \rightarrow p} \frac{f(x) - f(p)}{\sqrt{x} - \sqrt{p}}. \][/tex]

We recognize that [tex]\( f(x) = x^{3/2} \)[/tex], so we need to evaluate this difference quotient as [tex]\( x \)[/tex] approaches [tex]\( p \)[/tex]. Let's use L'Hopital's Rule since we have the indeterminate form [tex]\( \frac{0}{0} \)[/tex].

By L'Hopital’s Rule, we need to compute the derivatives of the numerator and the denominator. The derivative of [tex]\( f(x) = x^{3/2} \)[/tex] is
[tex]\[ f'(x) = \frac{d}{dx}(x^{3/2}) = \frac{3}{2}x^{1/2}. \][/tex]
The derivative of [tex]\( g(x) = \sqrt{x} \)[/tex] (which is the denominator) is
[tex]\[ g'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2}x^{-1/2}. \][/tex]

Applying L'Hopital's Rule, we get
[tex]\[ \lim_{x \rightarrow p} \frac{f(x) - f(p)}{g(x) - g(p)} = \lim_{x \rightarrow p} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow p} \frac{\frac{3}{2}x^{1/2}}{\frac{1}{2} x^{-1/2}}. \][/tex]

Simplify the fraction:
[tex]\[ \frac{\frac{3}{2} x^{1/2}}{\frac{1}{2} x^{-1/2}} = \frac{3}{2} x^{1/2} \cdot \frac{2}{1} x^{1/2} = 3 x. \][/tex]

Finally, as [tex]\( x \)[/tex] approaches [tex]\( p \)[/tex], the expression simplifies to:
[tex]\[ 3x \quad \text{evaluated at} \quad x = p \quad \text{gives} \quad 3p. \][/tex]

Therefore, the limit is
[tex]\[ \boxed{3p}, \][/tex]
corresponding to option (B).