Question

How do you evaluate the limit $\dfrac{{{x}^{2}}-7x+10}{x-2}$ as x approaches $2$?

Answer 1

For solving this limit, we must first substitute $x=2$ into the given function. By substituting, we will find out the limit to be in the indeterminate form of $\dfrac{0}{0}$. For solving it, we need to factorise the numerator ${{x}^{2}}-7x+10$ by dividing it by its factor $\left( x-2 \right)$. Then finally, by cancelling the common factor $\left( x-2 \right)$ from the numerator and the denominator, we will obtain the required limit.

Complete step-by-step answer:
Let us consider the function of x given in the above question as
$\Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}-7x+10}{x-2}$
According to the above question, we have to evaluate the limit of $f\left( x \right)$ as x approaches $2$. Let us put $x=2$ in the above function so as to evaluate $f\left( 2 \right)$ as

& \Rightarrow f\left( 2 \right)=\dfrac{{{2}^{2}}-7\left( 2 \right)+10}{2-2} \\\ & \Rightarrow f\left( 2 \right)=\dfrac{4-14+10}{2-2} \\\ & \Rightarrow f\left( 2 \right)=\dfrac{-10+10}{2-2} \\\ & \Rightarrow f\left( 2 \right)=\dfrac{0}{0} \\\ \end{aligned}$$ From the above we can say that as x approaches $2$, both the numerator and the denominator will approach zero so that we will obtain the indeterminate form of $$\dfrac{0}{0}$$. In order to evaluate this limit, we try out factorizing the numerator, which is equal to ${{x}^{2}}-7x+10$. Since on putting $x=2$ the numerator gives zero, by factor theorem we can say that $\left( x-2 \right)$ is a factor of the numerator. Therefore we divide ${{x}^{2}}-7x+10$ by $\left( x-2 \right)$ as shown. $$x-2\overset{x-5}{\overline{\left){\begin{aligned} & {{x}^{2}}-7x+10 \\\ & \underline{{{x}^{2}}-2x} \\\ & -5x+10 \\\ & \underline{-5x+10} \\\ & \underline{0} \\\ \end{aligned}}\right.}}$$ From the above division we can write $\Rightarrow {{x}^{2}}-7x+10=\left( x-2 \right)\left( x-5 \right)$ Putting this in the given function, it becomes $\Rightarrow f\left( x \right)=\dfrac{\left( x-2 \right)\left( x-5 \right)}{x-2}$ Now considering the limits as x approaches $2$ we get $\Rightarrow \displaystyle \lim_{x \to 2}f\left( x \right)=\displaystyle \lim_{x \to 2}\dfrac{\left( x-2 \right)\left( x-5 \right)}{\left( x-2 \right)}$ Cancelling $\left( x-2 \right)$ from the numerator and the denominator, we get $\begin{aligned} & \Rightarrow \displaystyle \lim_{x \to 2}f\left( x \right)=\displaystyle \lim_{x \to 2}\left( x-5 \right) \\\ & \Rightarrow \displaystyle \lim_{x \to 2}f\left( x \right)=2-5 \\\ & \Rightarrow \displaystyle \lim_{x \to 2}f\left( x \right)=-3 \\\ \end{aligned}$ Hence, the required limit is equal to $-3$. **Note:** For solving the limits which are of the indeterminate form of $$\dfrac{0}{0}$$, we can also use the L hospital’s rule. For this, we need to differentiate the numerator and the denominator separately, and then evaluate the limit of the resulting fraction. Also, before evaluating a limit, we must check whether it is a determinate or an indeterminate limit. This is because the determinate limits requires just the substitution for the evaluation.