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Question: How do you find the limit of \(\dfrac{{{x^3} - 27}}{{{x^2} - 9}}\) as \(x\) approaches to \(3\)?...

How do you find the limit of x3−27x2−9\dfrac{{{x^3} - 27}}{{{x^2} - 9}} as xx approaches to 33?

Explanation

Solution

In this problem we have given a reciprocal term with the variable xx. And we are asked to find the limit of the given term as xx approaches to some integer.
If we directly substitute the limit to the given term as xx approaches a given integer, then there is a chance for the given term becoming zero. So we use some expansions to get the required answer.

Formula used: We need to remember two formulas to solve this problem, they are
a2−b2=(a−b)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)
a3−b3=(a−b)(a2+b2+ab){a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)

Complete step-by-step solution:
We have given the expression x3−27x2−9\dfrac{{{x^3} - 27}}{{{x^2} - 9}}
We can rewrite the given expression by using the formulas which we noted above,
⇒x3−27x2−9=(x−3)(x2+3x+9)(x−3)(x+3)−−−−−(1)\Rightarrow \dfrac{{{x^3} - 27}}{{{x^2} - 9}} = \dfrac{{\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - - - - - (1)
As long as x≠3x \ne 3, which is true as long as xx is only approaches to33
The factors (x−3)\left( {x - 3} \right)can be cancelled out of the numerator and denominator, we get
⇒x3−27x2−9=(x2+3x+9)(x+3)\Rightarrow \dfrac{{{x^3} - 27}}{{{x^2} - 9}} = \dfrac{{\left( {{x^2} + 3x + 9} \right)}}{{\left( {x + 3} \right)}}
Now let’s apply the limit to the given expression, implies
⇒lim⁡x→3x3−27x2−9=lim⁡x→3(x2+3x+9)x+3\Rightarrow \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^3} - 27}}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to 3} \dfrac{{\left( {{x^2} + 3x + 9} \right)}}{{x + 3}}
⇒lim⁡x→3x2+3x+9x+3\Rightarrow \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} + 3x + 9}}{{x + 3}}
Now replace xx by 33 in the above equation, we get
⇒lim⁡x→332+3.3+93+3\Rightarrow \mathop {\lim }\limits_{x \to 3} \dfrac{{{3^2} + 3.3 + 9}}{{3 + 3}}
⇒9+9+96−−−−−(2)\Rightarrow \dfrac{{9 + 9 + 9}}{6} - - - - - (2), calculating the numbers, we get
⇒276\Rightarrow \dfrac{{27}}{6} Cancelling the numbers in the numerator and denominator, we get
⇒92\Rightarrow \dfrac{9}{2}
Therefore, when limit xx approaches to 33 in the expression x3−27x2−9\dfrac{{{x^3} - 27}}{{{x^2} - 9}}, the answer will be 92\dfrac{9}{2}.

Hence the required answer is 92\dfrac{9}{2}.

Note: In equation (1) the factors (x−3)\left( {x - 3} \right)can be cancelled out of the numerator and denominator since xx is not equal to 33. And also after the equation (1) we applied limit xx approaches to 33. So the cancellation work became easy for us. And in equation (2) onwards we stopped applying the limit, since the limit of a constant is the constant itself.