Question

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

Answer 1

Hint: To solve this question it is important to use the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ at first step. After this we will get the respective values of x for equation ${{x}^{2}}-3x$. By this we will have the required factor of it. After this the substitution process will start in which we will substitute x as 3 to get the desired result.

Complete step-by-step answer:
We will start solving this question by first understanding the concept of a limit. By the term limit we mean the closeness of the given function to the given limit. According to this question, we need to find the limit of the function $-\dfrac{{{x}^{2}}-3x}{x-3}$ as x comes closer to the point 3. To solve this function we will first simplify it just by using factorization. Factorization is a method in which we separate a quadratic equation into two simpler equations by taking solutions of that particular function. For example consider the equation \[{{x}^{2}}-3x\]. The solution of this function will be the one which will lead it to 0. Using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and taking a = 1, b = – 3. Therefore,
$\begin{align}
  & x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( 0 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{3\pm \sqrt{9}}{2} \\
 & \Rightarrow x=\dfrac{3\pm 3}{2} \\
 & \Rightarrow x=\dfrac{6}{2},\dfrac{0}{2} \\
 & \Rightarrow x=3,0 \\
\end{align}$
Thus, we can now reduce the equation \[{{x}^{2}}-3x\] into the factors \[\left( x-3 \right)\left( x-0 \right)=\left( x-3 \right)x\].
Now, after this step we are going to evaluate the function
\[\begin{align}
  & \displaystyle \lim_{x \to 3}\left( -\dfrac{{{x}^{2}}-3}{x-3} \right)=\displaystyle \lim_{x \to 3}\left( -\dfrac{x\left( x-3 \right)}{x-3} \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 3}\left( -\dfrac{{{x}^{2}}-3}{x-3} \right)=\displaystyle \lim_{x \to 3}\left( -x \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 3}\left( -\dfrac{{{x}^{2}}-3}{x-3} \right)=-3 \\
\end{align}\]
Hence, the correct limit of the function given to us is – 3.

Note:
This question can also be solved by direct factorization of the equation ${{x}^{2}}-3x$. Take the following process as an example,
$\begin{align}
  & {{x}^{2}}-3\,x=0 \\
 & \Rightarrow x\left( x-3 \right)=0 \\
\end{align}$
We cannot substitute the value of x as 3 at first. This is due to the fact that the value of given function will be reduced to $\displaystyle \lim_{x \to 3}\left( -\dfrac{{{x}^{2}}-3x}{x-3} \right)=\displaystyle \lim_{x \to 3}\left( -\dfrac{{{\left( 3 \right)}^{2}}-3\left( 3 \right)}{3-3} \right)$ resulting into the denominator getting undefined. So, to restrict this problem we need to simplify the function first and then substitute the value x as 3.