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Question: How do you evaluate the limit \(-\dfrac{{{x}^{2}}-3x}{x-3}\) as x approaches 3?...

How do you evaluate the limit −x2−3xx−3-\dfrac{{{x}^{2}}-3x}{x-3} as x approaches 3?

Explanation

Solution

To solve this question it is important to use the formula x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} at first step. After this we will get the respective values of x for equation x2−3x{{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 −x2−3xx−3-\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 x2−3x{{x}^{2}}-3x. The solution of this function will be the one which will lead it to 0. Using the formula x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} and taking a = 1, b = – 3. Therefore,
x=−(−3)±(−3)2−4(1)(0)2(1) ⇒x=3±92 ⇒x=3±32 ⇒x=62,02 ⇒x=3,0 \begin{aligned} & 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{aligned}
Thus, we can now reduce the equation x2−3x{{x}^{2}}-3x into the factors (x−3)(x−0)=(x−3)x\left( x-3 \right)\left( x-0 \right)=\left( x-3 \right)x.
Now, after this step we are going to evaluate the function

& \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{aligned}$$ 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, $\begin{aligned} & {{x}^{2}}-3\,x=0 \\\ & \Rightarrow x\left( x-3 \right)=0 \\\ \end{aligned}$ 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.