Question

What is the limit of $\dfrac{{{x}^{3}}-8}{x-2}$ as x approaches 2?

Answer 1

Hint: We first take the factorisation of the given polynomial ${{x}^{3}}-8$ in the numerator according to the identity ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. The division cancels out the polynomial $\left( x-2 \right)$ in the factorisation and gives $\left( {{x}^{2}}+2x+4 \right)$ as the quotient. We also verify the result with the long division.

Complete step by step answer:
We need to find the value of the limit of $\dfrac{{{x}^{3}}-8}{x-2}$ as x approaches 2. The mathematical form is
$\displaystyle \lim_{x \to 2}\dfrac{{{x}^{3}}-8}{x-2}$.
We have been given a fraction whose denominator and numerator both are polynomial.
We need to find the value of the quotient.
The numerator is a cubic expression. It’s a difference of two cube numbers. We factorise the given difference of the cubes according to the identity ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$.
We have ${{x}^{3}}-8$ in the numerator which gives ${{x}^{3}}-8={{x}^{3}}-{{2}^{3}}$.
We put the values $a=x;b=2$.
We get ${{x}^{3}}-{{2}^{3}}=\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$.
We can see the term ${{x}^{3}}-8$ is a multiplication of two polynomials $\left( x-2 \right)$ and $\left( {{x}^{2}}+2x+4 \right)$.
Therefore, \[\dfrac{{{x}^{3}}-8}{x-2}=\dfrac{\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)}{\left( x-2 \right)}=\left( {{x}^{2}}+2x+4 \right)\].
So, $\displaystyle \lim_{x \to 2}\dfrac{{{x}^{3}}-8}{x-2}=\displaystyle \lim_{x \to 2}\left( {{x}^{2}}+2x+4 \right)$.
Putting the limit value, we get $\displaystyle \lim_{x \to 2}\dfrac{{{x}^{3}}-8}{x-2}=\left( {{2}^{2}}+2\times 2+4 \right)=12$

The limit of $\dfrac{{{x}^{3}}-8}{x-2}$ as x approaches 2 is 12.

Note: The precise definition of a limit is something we use as a proof for the existence of a limit. When we’re evaluating a limit, we’re looking at the function as it approaches a specific point. we approach a particular value of x, the function itself gets closer and closer to a particular value.