Question

i) $lim_{x
ightarrow1}f_1(x)$ where $f_1(x)=\begin{cases}\frac{x^{2}-4x - 5}{1 - x^{2}}& \text{if }xgeq1\\frac{x^{2}+x - 2}{1 - x^{2}}& \text{if }x<1end{cases}$

Explanation:

Step1: Find left - hand limit

We find $\lim_{x
ightarrow1^{-}}f_1(x)$. Since $x
ightarrow1^{-}$ (i.e., $x < 1$), $f_1(x)=\frac{x^{2}+x - 2}{1 - x^{2}}$. Factor the numerator and denominator: $x^{2}+x - 2=(x + 2)(x - 1)$ and $1 - x^{2}=(1 + x)(1 - x)$. Then $\lim_{x
ightarrow1^{-}}\frac{x^{2}+x - 2}{1 - x^{2}}=\lim_{x
ightarrow1^{-}}\frac{(x + 2)(x - 1)}{(1 + x)(1 - x)}=\lim_{x
ightarrow1^{-}}\frac{-(x + 2)(1 - x)}{(1 + x)(1 - x)}=\lim_{x
ightarrow1^{-}}\frac{-(x + 2)}{1 + x}$. Substitute $x = 1$: $\frac{-(1 + 2)}{1+1}=-\frac{3}{2}$.

Step2: Find right - hand limit

We find $\lim_{x
ightarrow1^{+}}f_1(x)$. Since $x
ightarrow1^{+}$ (i.e., $x\geq1$), $f_1(x)=\frac{x^{2}-4x - 5}{1 - x^{2}}$. Factor the numerator and denominator: $x^{2}-4x - 5=(x - 5)(x+1)$ and $1 - x^{2}=(1 + x)(1 - x)$. Then $\lim_{x
ightarrow1^{+}}\frac{x^{2}-4x - 5}{1 - x^{2}}=\lim_{x
ightarrow1^{+}}\frac{(x - 5)(x + 1)}{(1 + x)(1 - x)}=\lim_{x
ightarrow1^{+}}\frac{x - 5}{1 - x}$. Substitute $x = 1$, we get $\frac{1-5}{1 - 1}=\frac{-4}{0}$, which is undefined. But we can rewrite it as $\lim_{x
ightarrow1^{+}}\frac{x - 5}{1 - x}=\lim_{x
ightarrow1^{+}}\frac{-(5 - x)}{1 - x}$. As $x
ightarrow1^{+}$, the limit does not exist.

Since the left - hand limit $=-\frac{3}{2}$ and the right - hand limit does not exist, $\lim_{x
ightarrow1}f_1(x)$ does not exist.

Answer:

The limit $\lim_{x
ightarrow1}f_1(x)$ does not exist.