17. $f(x)=\frac{(x - 1)(x + 5)}{(x - 1)(x + 2)}$\nleft: $lim_{x\rightarrow1^{-}}f(x)=2$\nright…

Explanation:

Step1: Simplify the function

For $x\neq1$, we can cancel out the common factor $(x - 1)$ in the numerator and denominator of $f(x)=\frac{(x - 1)(x + 5)}{(x - 1)(x + 2)}$. So $f(x)=\frac{x + 5}{x+2}$ for $x\neq1$.

Step2: Calculate the left - hand limit

To find $\lim_{x\rightarrow1^{-}}f(x)$, we substitute $x$ values approaching 1 from the left into $y = \frac{x + 5}{x+2}$. $\lim_{x\rightarrow1^{-}}\frac{x + 5}{x+2}=\frac{1 + 5}{1+2}=\frac{6}{3}=2$.

Step3: Calculate the right - hand limit

To find $\lim_{x\rightarrow1^{+}}f(x)$, we substitute $x$ values approaching 1 from the right into $y=\frac{x + 5}{x+2}$. $\lim_{x\rightarrow1^{+}}\frac{x + 5}{x+2}=\frac{1 + 5}{1+2}=\frac{6}{3}=2$.

Answer:

The left - hand limit $\lim_{x\rightarrow1^{-}}f(x)=2$ and the right - hand limit $\lim_{x\rightarrow1^{+}}f(x)=2$.