Question

How do you find the limit of $\dfrac{{{x^3} - 27}}{{{x^2} - 9}}$ as $x$ approaches to $3$?

Answer 1

Hint: In this problem we have given a reciprocal term with the variable $x$. And we are asked to find the limit of the given term as $x$ approaches to some integer.
If we directly substitute the limit to the given term as $x$ 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
${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$

Complete step-by-step solution:
We have given the expression $\dfrac{{{x^3} - 27}}{{{x^2} - 9}}$
We can rewrite the given expression by using the formulas which we noted above,
\[ \Rightarrow \dfrac{{{x^3} - 27}}{{{x^2} - 9}} = \dfrac{{\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - - - - - (1)\]
As long as $x \ne 3$, which is true as long as $x$ is only approaches to$3$
The factors $\left( {x - 3} \right)$can be cancelled out of the numerator and denominator, we get
$ \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
$ \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}}$
$ \Rightarrow \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} + 3x + 9}}{{x + 3}}$
Now replace $x$ by $3$ in the above equation, we get
$ \Rightarrow \mathop {\lim }\limits_{x \to 3} \dfrac{{{3^2} + 3.3 + 9}}{{3 + 3}}$
$ \Rightarrow \dfrac{{9 + 9 + 9}}{6} - - - - - (2)$, calculating the numbers, we get
$ \Rightarrow \dfrac{{27}}{6}$ Cancelling the numbers in the numerator and denominator, we get
$ \Rightarrow \dfrac{9}{2}$
Therefore, when limit $x$ approaches to $3$ in the expression $\dfrac{{{x^3} - 27}}{{{x^2} - 9}}$, the answer will be $\dfrac{9}{2}$.

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

Note: In equation (1) the factors $\left( {x - 3} \right)$can be cancelled out of the numerator and denominator since $x$ is not equal to $3$. And also after the equation (1) we applied limit $x$ approaches to $3$. 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.