find the limit as x approaches 1 for the following rational function (y value)\n19. $f(x)=\frac{-2(x + 3)(x…

find the limit as x approaches 1 for the following rational function (y value)\n19. $f(x)=\frac{-2(x + 3)(x + 1)}{(x - 1)^2}$\nleft:\nright:

find the limit as x approaches 1 for the following rational function (y value)\n19. $f(x)=\frac{-2(x + 3)(x + 1)}{(x - 1)^2}$\nleft:\nright:

Answer

Explanation:

Step1: Analyze left - hand limit

As (x\to1^{-}), we consider values of (x) that are less than 1 but getting closer to 1. The numerator (-2(x + 3)(x + 1)) evaluated at (x = 1) is (-2(1 + 3)(1+1)=-2\times4\times2=-16). The denominator ((x - 1)^2>0) for (x\neq1). As (x\to1^{-}), ((x - 1)^2\to0^{+}). So, (\lim_{x\to1^{-}}\frac{-2(x + 3)(x + 1)}{(x - 1)^2}=-\infty).

Step2: Analyze right - hand limit

As (x\to1^{+}), we consider values of (x) that are greater than 1 but getting closer to 1. The numerator (-2(x + 3)(x + 1)) is still (-16) when evaluated at (x = 1). The denominator ((x - 1)^2>0) for (x\neq1). As (x\to1^{+}), ((x - 1)^2\to0^{+}). So, (\lim_{x\to1^{+}}\frac{-2(x + 3)(x + 1)}{(x - 1)^2}=-\infty).

Answer:

Left: (-\infty) Right: (-\infty)