The bases of the terms in numerators are $x+2$ and $a+2$, the binomial $x+2$ is a variable expression and the binomial $a+2$ is a constant expression but the difference between them is equal to the denominator. Each binomial has $\dfrac{5}{3}$ as its exponent.
$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{x-a}}$
The limit of this algebraic function is similar to the limit of $\dfrac{x^n-a^n}{x-a}$ as $x$ approaches $a$ formula. So, let’s try to convert this algebraic function in this form of this limit rule.
In this limit problem, the difference of the binomials in the terms of the numerator is exactly equals to the function in the denominator. So, Add and subtract number $2$ in the denominator for making the function in the denominator similar to the numerator.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{x+2-2-a}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{x+2-a-2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{(x+2)-(a+2)}}$
If $x \to a$, then $x+2 \to a+2$. It states that if $x$ approaches $a$ then $x+2$ tends to $a+2$.
$= \,\,\,$ $\displaystyle \large \lim_{x+2 \,\to\, a+2}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{(x+2)-(a+2)}}$
According to limit of (xn-an)/(x-a) as x approaches a standard rule, the limit of the algebraic function can be evaluated mathematically.
$= \,\,\,$ $\dfrac{5}{3}{(a+2)}^{\frac{5}{3}-1}$
$= \,\,\,$ $\dfrac{5}{3}{(a+2)}^{\frac{5-1 \times 3}{3}}$
$= \,\,\,$ $\dfrac{5}{3}{(a+2)}^{\frac{5-3}{3}}$
$= \,\,\,$ $\dfrac{5}{3}{(a+2)}^{\frac{2}{3}}$
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