Question

Find out the limit for $\dfrac{{3x + 9}}{{{x^2} - 9}}$ as $x$ approaches $- 3$.

Answer 1

The given question deals with finding out the limit for the given function as $x$ approaches to $- 3$. In order to find out the limit, we will at first factorize the denominator. After which we will eliminate the common factor. Ultimately, we will apply the limit at $x = - 3$ to find the answer.

Complete step by step answer:
We have,
$\Rightarrow \mathop {\lim }\limits_{x \to - 3} \dfrac{{3x + 9}}{{{x^2} - 9}}$
Here, we factorize the numerator by taking out the common factor.
Thus, we get
$\Rightarrow \mathop {\lim }\limits_{x \to - 3} \dfrac{{3\left( {x + 3} \right)}}{{\left( {{x^2} - 9} \right)}} - - - - - \left( 1 \right)$
Now, we factorize the denominator using the algebraic identity of difference of squares.
The identity of difference of squares is, $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$
Here, we rewrite our given function as,
$\Rightarrow \mathop {\lim }\limits_{x \to - 3} \dfrac{{3\left( {x + 3} \right)}}{{\left( {{x^2} - {3^2}} \right)}}$
Therefore, after using the identity of difference of squares, the above function becomes,
$\Rightarrow \mathop {\lim }\limits_{x \to - 3} \dfrac{{3\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}$
Now, after eliminating common factors from the above function, it becomes,
$\Rightarrow \mathop {\lim }\limits_{x \to - 3} \dfrac{3}{{\left( {x - 3} \right)}}$
Now, we apply the limit. Therefore, we put$x = - 3$

Thus we get,
$\dfrac{3}{{ - 3 - 3}} = \dfrac{3}{{ - 6}} = - \dfrac{3}{6} = - \dfrac{1}{2}$ Which is our required limit.

Note: One of the important things to note here is that other than the standard algebraic identities used to solve limits, there are five more very important trigonometric properties of limit. These other identities are useful in solving almost every question that involves limits. They are as follows:
Let a, k and P, Q represent real numbers and $f$ and $g$ be functions. Such that, $\mathop {\lim }\limits_{x \to a} f\left( x \right)$and$\mathop {\lim }\limits_{x \to a} g\left( x \right)$are limits that exist and are finite. The properties for such limits are as follows:
For a constant say, k it is: $\mathop {\lim }\limits_{x \to a} k = k$
Constant times a function is, $\mathop {\lim }\limits_{x \to a} k.f\left( x \right) = k\mathop {\lim }\limits_{x \to a} f\left( x \right) = ka$
For addition and subtraction of function, the properties are$\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right) = P + Q$ and $\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right) = P - Q$respectively.
Last but not the least, for division and multiplication of functions, the properties are, $\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right).g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) \times \mathop {\lim }\limits_{x \to a} g\left( x \right)$ and $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}$ respectively.