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Question: What is the limit of \(\dfrac{{{x}^{3}}-8}{x-2}\) as x approaches 2?...

What is the limit of x3−8x−2\dfrac{{{x}^{3}}-8}{x-2} as x approaches 2?

Explanation

Solution

We first take the factorisation of the given polynomial x3−8{{x}^{3}}-8 in the numerator according to the identity a3−b3=(a−b)(a2+ab+b2){{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right). The division cancels out the polynomial (x−2)\left( x-2 \right) in the factorisation and gives (x2+2x+4)\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 x3−8x−2\dfrac{{{x}^{3}}-8}{x-2} as x approaches 2. The mathematical form is
lim⁡x→2x3−8x−2\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 a3−b3=(a−b)(a2+ab+b2){{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right).
We have x3−8{{x}^{3}}-8 in the numerator which gives x3−8=x3−23{{x}^{3}}-8={{x}^{3}}-{{2}^{3}}.
We put the values a=x;b=2a=x;b=2.
We get x3−23=(x−2)(x2+2x+4){{x}^{3}}-{{2}^{3}}=\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right).
We can see the term x3−8{{x}^{3}}-8 is a multiplication of two polynomials (x−2)\left( x-2 \right) and (x2+2x+4)\left( {{x}^{2}}+2x+4 \right).
Therefore, x3−8x−2=(x−2)(x2+2x+4)(x−2)=(x2+2x+4)\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, lim⁡x→2x3−8x−2=lim⁡x→2(x2+2x+4)\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 lim⁡x→2x3−8x−2=(22+2×2+4)=12\displaystyle \lim_{x \to 2}\dfrac{{{x}^{3}}-8}{x-2}=\left( {{2}^{2}}+2\times 2+4 \right)=12

The limit of x3−8x−2\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.