In this limits problem, $x$ is a variable and $a$ is a constant. The limit of the algebraic function $\dfrac{x\sqrt{x}-a\sqrt{a}}{x-a}$ has to be calculated as $x$ approaches $a$.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x\sqrt{x}-a\sqrt{a}}{x-a}}$

The square root symbol represents an exponent of $\dfrac{1}{2}$ mathematically. So, express square root in its actual form.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x \times x^{\frac{1}{2}}-a \times a^{\frac{1}{2}}}{x-a}}$

In numerator, the bases of factors in each term is same. So, it can be simplified by using product rule of exponents.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{(1+\frac{1}{2})}-a^{(1+\frac{1}{2})}}{x-a}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{\frac{3}{2}}-a^{\frac{3}{2}}}{x-a}}$

$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^{\frac{3}{2}}-a^{\frac{3}{2}}}{x-a}}$

The limit of this algebraic function is same as the limit formula for the algebraic function $\dfrac{x^n-a^n}{x-a}$ as $x$ approaches $a$ formula. As per this limit property, the limit of this algebraic function can be evaluated.

$= \,\,\,$ $\dfrac{3}{2}a^{(\frac{3}{2}-1)}$

$= \,\,\,$ $\dfrac{3}{2}a^{(\frac{3-2}{2})}$

$= \,\,\,$ $\dfrac{3}{2}a^{\frac{1}{2}}$

$= \,\,\,$ $\dfrac{3}{2}\sqrt{a}$

Therefore, it’s evaluated that the limit of the algebraic function $\dfrac{x\sqrt{x}-a\sqrt{a}}{x-a}$ as $x$ approaches $a$ is equal to $\dfrac{3}{2}\sqrt{a}$.

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