Question

i) $lim_{x \to 1} f_2(x)$ where $f_2(x)=\begin{cases}\frac{2 - x - x^2}{1 - x^2}&\text{if }xgeq1\\frac{\frac{1}{sqrt{2 - x}}-\frac{1}{x}}{x - 1}&\text{if }x < 1end{cases}$

Explanation:

Step1: Calculate right - hand limit

For $x\geq1$, $f_2(x)=\frac{2 - x - x^2}{1 - x^2}$. Factor the numerator and denominator: $2 - x - x^2=-(x^2 + x - 2)=-(x + 2)(x - 1)$ and $1 - x^2=(1 + x)(1 - x)$. Then $\lim_{x
ightarrow1^+}f_2(x)=\lim_{x
ightarrow1^+}\frac{-(x + 2)(x - 1)}{(1 + x)(1 - x)}=\lim_{x
ightarrow1^+}\frac{(x + 2)}{(1 + x)}=\frac{1+2}{1 + 1}=\frac{3}{2}$.

Step2: Calculate left - hand limit

For $x<1$, $f_2(x)=\frac{\frac{1}{\sqrt{2 - x}}-\frac{1}{x}}{x - 1}$. First, find a common denominator for the numerator: $\frac{x-\sqrt{2 - x}}{x\sqrt{2 - x}}$. Then rationalize the numerator: $\frac{(x-\sqrt{2 - x})(x+\sqrt{2 - x})}{(x - 1)x\sqrt{2 - x}(x+\sqrt{2 - x})}=\frac{x^2-(2 - x)}{(x - 1)x\sqrt{2 - x}(x+\sqrt{2 - x})}=\frac{x^2+x - 2}{(x - 1)x\sqrt{2 - x}(x+\sqrt{2 - x})}$. Factor the numerator: $x^2+x - 2=(x + 2)(x - 1)$. So $\lim_{x
ightarrow1^-}f_2(x)=\lim_{x
ightarrow1^-}\frac{(x + 2)}{x\sqrt{2 - x}(x+\sqrt{2 - x})}=\frac{1+2}{1\times\sqrt{2 - 1}(1+\sqrt{2 - 1})}=\frac{3}{2}$.

Answer:

$\frac{3}{2}$