Math Doubts

Evaluate $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{x-a}}$

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.

Convert the denominator similar to numerator

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.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{x+2-2-a}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{x+2-a-2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{(x+2)-(a+2)}}$

Set the Input of the Limit function

If $x \to a$, then $x+2 \to a+2$. It states that if $x$ approaches $a$ then $x+2$ tends to $a+2$.

$= \,\,\,$ $\displaystyle \large \lim_{x+2 \,\to\, a+2}{\normalsize \dfrac{{(x+2)}^\frac{5}{3}-{(a+2)}^\frac{5}{3}}{(x+2)-(a+2)}}$

Evaluate Limit of Algebraic function

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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