Math Doubts

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.

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

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

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

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

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

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

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the math problems in different methods with understandable steps and worksheets on every concept for your practice.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved