Question

ii) $lim_{x \to 1} f_2(x)$ 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 left - hand limit

We need to find $\lim_{x
ightarrow1^{-}}f_2(x)$. First, simplify $\frac{\frac{1}{\sqrt{2 - x}}-\frac{1}{x}}{x - 1}$.
\[

$$\begin{align*} \frac{\frac{1}{\sqrt{2 - x}}-\frac{1}{x}}{x - 1}&=\frac{\frac{x-\sqrt{2 - x}}{x\sqrt{2 - x}}}{x - 1}\\ &=\frac{x-\sqrt{2 - x}}{(x - 1)x\sqrt{2 - x}}\\ &=\frac{(x-\sqrt{2 - x})(x+\sqrt{2 - x})}{(x - 1)x\sqrt{2 - x}(x+\sqrt{2 - x})}\\ &=\frac{x^{2}-(2 - x)}{(x - 1)x\sqrt{2 - x}(x+\sqrt{2 - x})}\\ &=\frac{x^{2}+x - 2}{(x - 1)x\sqrt{2 - x}(x+\sqrt{2 - x})}\\ &=\frac{(x - 1)(x + 2)}{(x - 1)x\sqrt{2 - x}(x+\sqrt{2 - x})}\\ &=\frac{x + 2}{x\sqrt{2 - x}(x+\sqrt{2 - x})} \end{align*}$$

\]
Now, $\lim_{x
ightarrow1^{-}}\frac{x + 2}{x\sqrt{2 - x}(x+\sqrt{2 - x})}=\frac{1+2}{1\times\sqrt{2 - 1}(1+\sqrt{2 - 1})}=\frac{3}{1\times1\times(1 + 1)}=\frac{3}{2}$.

Step2: Calculate right - hand limit

We need to find $\lim_{x
ightarrow1^{+}}f_2(x)$. For $x\geq1$, $f_2(x)=\frac{2 - x - x^{2}}{1 - x^{2}}=\frac{-(x^{2}+x - 2)}{-(x^{2}-1)}=\frac{(x + 2)(x - 1)}{(x + 1)(x - 1)}=\frac{x + 2}{x + 1}$ (for $x
eq1$).
Then $\lim_{x
ightarrow1^{+}}\frac{x + 2}{x + 1}=\frac{1+2}{1+1}=\frac{3}{2}$.

Answer:

$\frac{3}{2}$