Question

How do you find the limit of \[\dfrac{{{x^2} + x - 6}}{{{x^2} - 9}}\,as\,x \to - 3\]

Answer 1

Hint: To solve the question of limit we need to know that limit is the closest value for the given number, in order to find the value for a given function at the limit value, we need to put the value given in the function, and then on solving we can get the answer that is the limit value for the function at the given value.

Complete step by step solution:
Here the given question is to find the limit value of the function given, at the value of minus three, here in order to solve the question we need to put the value given in the function and then on simplifying the given expression we can get the solution required, on solving we get:
In the given equation here we are going to put the limit sign along with the given value, we get:
 \[ \Rightarrow \mathop {\lim }\limits_{x \to - 3} \dfrac{{{x^2} + x - 6}}{{{x^2} - 9}}\]
Now putting the limiting value in the expression we get:
 \[ \Rightarrow \mathop {\lim }\limits_{x \to - 3} \dfrac{{{{( - 3)}^2} + ( - 3) - 6}}{{{{( - 3)}^2} - 9}}\]
On solving the expression we get:
 \[ \Rightarrow \mathop {\lim }\limits_{x \to - 3} \dfrac{{9 - 3 - 6}}{{9 - 9}} = \dfrac{0}{0}\]
Here we get the form of \[\dfrac{0}{0}\] , hence we have to use the properties of limit which says whenever this sort of expression is simplified into \[\dfrac{0}{0}\] form then to get the value we need to differentiate the denominator as well as numerator separately and then put the value to get the answer:
 \[ \Rightarrow \dfrac{{\dfrac{d}{{dx}}({x^2} + x - 6)}}{{\dfrac{d}{{dx}}({x^2} - 9)}} = \dfrac{{2x + 1}}{{2x}}\]
Now putting limit sign with the value given we get:
 \[ \Rightarrow \mathop {\lim }\limits_{x \to - 3} \dfrac{{2x + 1}}{{2x}} = \dfrac{{2( - 3) + 1}}{{2( - 3)}} = \dfrac{{ - 6 + 1}}{{ - 6}} = \dfrac{{ - 5}}{{ - 6}} = \dfrac{5}{6}\]
Here we got the answer for the given question.
So, the correct answer is “$\dfrac{5}{6}$”.

Note: Limit question can be solved by directly putting the limit value, but sometimes the expression given is of indeterminate form hence we need to solve it by using limit properties, the properties needed to be chosen depends upon the expression we get to solve.