Question

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

Answer 1

Hint:Determining the limits algebraically can be in many ways. In case of polynomials, we can obtain the limits by simply substituting the limiting value of x into the polynomials. Sometimes we use some kind of radical conjugate. In the type of questions as given above it is more advisable to proceed by making the factors of the equation. After simplifying the equation, put the value of x to get the final answer.

Complete step-by-step solution-
In the given question, we have to find the limit of $\dfrac{{{x^2} - 9}}{{x - 3}}$when x approaches to 3.
That is , $\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{{x^2} - 9}}{{x - 3}}} \right)$ The limiting condition of x is 3 which is a constant number but if you notice carefully , if we substitute 3 in the equation then,
$ \Rightarrow \left( {\dfrac{{9 - 9}}{{3 - 3}}} \right) = \dfrac{0}{0}$
Therefore we have a ratio of two zeroes $\dfrac{0}{0}$ means that we cannot directly substitute the value of x.
In this case let us first factorise the equation,
Using the property, $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$
We get,
$\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{\left( {x + 3} \right)\left( {x - 3} \right)}}{{\left( {x - 3} \right)}}} \right)$
On further solving we get,
$\mathop {\lim }\limits_{x \to 3} \left( {x + 3} \right)$
Now we have exited the $\dfrac{0}{0}$ form hence now it is safe to substitute the value of x=3
$ \Rightarrow \mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{{x^2} - 9}}{{x - 3}}} \right) = 6$
The answer comes out to be 6 in this case also.
Note- The above question can also be solved by using L’ Hospital’s rule as it is of the form $\dfrac{0}{0}$ To apply L’ Hospital’s rule , differentiate both numerator and denominator with respect to x.

After differentiating we get,
$\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{2x}}{1}} \right)$ then substitute the value of x=3
Hence you can use this method to cross-check your answer whenever your equation is of the form $\dfrac{0}{0}$or $\dfrac{\infty }{\infty }$