Question

How do you evaluate the limit $\dfrac{{{x}^{2}}-5x+4}{{{x}^{2}}-2x-8}$ as x approaches 4?

Answer 1

Hint: We are going to take the help of the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the respective values of x in each of the equations ${{x}^{2}}-5x+4\,,\,{{x}^{2}}-2x-8$. This will change these equations into factors. Then, we will use these factors to solve the question further. After this simplification we will substitute the value of x as 4 to get the answer.

Complete step-by-step answer:
A limit basically means that how close is the given function close to the given limit. In reference to this question, we need to find the limit of the function \[\dfrac{{{x}^{2}}-5x+4}{{{x}^{2}}-2x-8}\] as x come closer to the point 4. 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 \[{{x}^{2}}-5x+4\]. 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 = - 5, c = 4. Therefore,
$\begin{align}
  & x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{25-16}}{2} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{9}}{2} \\
 & \Rightarrow x=\dfrac{5\pm 3}{2} \\
 & \Rightarrow x=\dfrac{8}{2},\dfrac{2}{2} \\
 & \Rightarrow x=4,1 \\
\end{align}$
Thus, we can now reduce the equation \[{{x}^{2}}-5x+4\] into the factors \[\left( x-1 \right)\left( x-4 \right)\]. Similarly, for \[{{x}^{2}}-2x-8\] we have,
$\begin{align}
  & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -8 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{2\pm \sqrt{4+32}}{2}=\dfrac{2\pm \sqrt{36}}{2} \\
 & \Rightarrow x=\dfrac{2\pm 6}{2} \\
 & \Rightarrow x=\dfrac{8}{2},\dfrac{-4}{2}=4,-2 \\
\end{align}$
So, we can have the factors of the equation \[{{x}^{2}}-2x-8\] as $\left( x-4 \right)\left( x+2 \right)$.
Now, after this step we are going to evaluate the function \[\displaystyle \lim_{x \to 4}\left( \dfrac{{{x}^{2}}-5x+4}{{{x}^{2}}-2x-8} \right)\] as the limit of x approaches to the number 4. We can do this by the following process,
\[\begin{align}
  & \displaystyle \lim_{x \to 4}\left( \dfrac{{{x}^{2}}-5x+4}{{{x}^{2}}-2x-8} \right)=\displaystyle \lim_{x \to 4}\dfrac{\left( x-4 \right)\left( x-1 \right)}{\left( x-4 \right)\left( x+2 \right)} \\
 & \Rightarrow \displaystyle \lim_{x \to 4}\left( \dfrac{{{x}^{2}}-5x+4}{{{x}^{2}}-2x-8} \right)=\displaystyle \lim_{x \to 4}\left( \dfrac{x-1}{x+2} \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 4}\left( \dfrac{{{x}^{2}}-5x+4}{{{x}^{2}}-2x-8} \right)=\dfrac{4-1}{4+2}=\dfrac{3}{6}=\dfrac{1}{2} \\
\end{align}\]
Hence, the correct limit of the function given to us is $\dfrac{1}{2}$.

Note:
We can also factor the equations ${{x}^{2}}-5x+4\,,\,{{x}^{2}}-2x-8$ directly. For example,
$\begin{align}
  & {{x}^{2}}-5x+4\,=0 \\
 & \Rightarrow {{x}^{2}}-4x-1x+4=0 \\
 & \Rightarrow x\left( x-4 \right)-1\left( x-4 \right)=0 \\
 & \Rightarrow \left( x-1 \right)\left( x-4 \right)=0 \\
\end{align}$
Similarly, we can factorize the equation ${{x}^{2}}-2x-8$. We cannot substitute the value of x as 4 at first. This is due to the fact that the value of given function will be reduced to $\displaystyle \lim_{x \to 4}\left( \dfrac{{{x}^{2}}-5x+4}{{{x}^{2}}-2x-8} \right)=\dfrac{{{\left( 4 \right)}^{2}}-5\left( 4 \right)+4}{{{\left( 4 \right)}^{2}}-2\left( 4 \right)-8}=\dfrac{16-20+4}{16-8-8}$ where the denominator gets undefined. So, to restrict this problem we need to simplify the function first and then substitute the value x as 4.