Evaluate $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3-8}{x^2-4}}$ by factorization

The limit of x cube minus 8 by x square minus 4 as x approaches 2 is indeterminate. The limit of the given rational function is indeterminate due to a factor $x-2$ in the denominator. So, if the factor $x-2$ is eliminated from the denominator, then it clears the route to find the limit of the given function.

$\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3-8}{x^2-4}}$

Now, let us learn how to find the limit of the $x$ cubed minus $8$ by $x$ squared minus $4$ as the value of $x$ is closer to $2$ by the factorisation.

Factorize the difference of cubes

Look at the numerator, $x^3-8$ is an expression. It can be written as a difference of two cubes.

$\implies$ $x^3-8$ $\,=\,$ $x^3-2^3$

According to the factorization by the difference of cubes, the difference of cubes can be written as a product of two factors by using the difference of cubes formula.

$a^3-b^3$ $\,=\,$ $(a-b)(a^2+b^2+ab)$

Now, take $a \,=\, x$ and $b \,=\, 2$, and substitute in the above algebraic identity.

$\implies$ $x^3-2^3$ $\,=\,$ $(x-2)(x^2+2^2+x \times 2)$

$\,\,\,\therefore\,\,\,\,\,\,$ $x^3-8$ $\,=\,$ $(x-2)(x^2+4+2x)$

Factorise the difference of squares

Consider the denominator and $x^2-4$ is an expression. It can be expressed as a difference of two squares.

$\implies$ $x^2-4$ $\,=\,$ $x^2-2^2$

The difference of squares can be factored by using the factorization formula for difference of squares.

$a^2-b^2$ $\,=\,$ $(a-b)(a+b)$

Suppose, $a \,=\, x$ and $b \,=\, 2$. Substitute them in the above formula to covert the difference of squares into factor form.

$\implies$ $x^2-2^2$ $\,=\,$ $(x-2)(x+2)$

$\,\,\,\therefore\,\,\,\,\,\,$ $x^2-4$ $\,=\,$ $(x-2)(x+2)$

Evaluate the Limit by direct substitution

We have successfully factored both algebraic expressions as follows.

$(1).\,\,\,$ $x^3-8$ $\,=\,$ $(x-2)(x^2+4+2x)$

$(2).\,\,\,$ $x^2-4$ $\,=\,$ $(x-2)(x+2)$

It is right time to replace the values of both difference of cubes and difference of squares in factoring form for finding the limit of the given rational function.

$\implies$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3-8}{x^2-4}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{(x-2)(x^2+4+2x)}{(x-2)(x+2)}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{(x-2) \times (x^2+4+2x)}{(x-2) \times (x+2)}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{\cancel{(x-2)} \times (x^2+4+2x)}{\cancel{(x-2)} \times (x+2)}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^2+4+2x}{x+2}}$

The simplification process has been completed and it is time to calculate the limit of the function as the value of $x$ is closer to $2$ by using the direct substitution method.

$\,\,\,=\,\,\,$ $\dfrac{2^2+4+2(2)}{2+2}$

$\,\,\,=\,\,\,$ $\dfrac{4+4+2 \times 2}{4}$

$\,\,\,=\,\,\,$ $\dfrac{4+4+4}{4}$

$\,\,\,=\,\,\,$ $\dfrac{12}{4}$

$\,\,\,=\,\,\,$ $\dfrac{\cancel{12}}{\cancel{4}}$

$\,\,\,=\,\,\,$ $3$