Question

find the limit as x approaches 1 for the following rational function (y value)

1. $f(x)=\frac{-2(x + 3)(x + 1)}{(x - 1)^2}$

left:
right:

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 \(x
eq1\). 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 \(x
eq1\). 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: \(-\infty\)