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• Notes

Let $F_{(x)}=\frac{x^{2}-x-2}{x-2}$

Check for continuity at $x=0$

(1) $f(a)=f(0)=\frac{0^{2}-0-2}{0-2}=\frac{-2}{-2}=1$

(2) $\lim _{x \rightarrow a} f(a)=\lim _{x \rightarrow 0} f(0)=\lim _{x \rightarrow 0} \frac{x^{2}-x-2}{x-2}=\frac{0-0-2}{0-2}= 1$

$f(0)=\lim _{x \rightarrow 0} f(x)$

\ at $x=0 \quad \longrightarrow \quad f$ is cont

Let $F(x) =\frac{|x-3|}{x-3}, \quad x \neq 3$

and $F(x) = 1$ at $x=3$

Check the Continuity at $x=3$

(1) $f(3)=1$

(2) $\lim _{x \rightarrow 3} \frac{|x-3|}{x-3}$

$\lim _{x \rightarrow 3^{+}} \frac{x-3}{x-3}=1$

$\lim _{x \rightarrow 3^{-}} \frac{-(x-3)}{(x-3)}=-1$

$\lim _{x \rightarrow 3^{+}} f(x) \neq \lim _{x \rightarrow 3^{-}} f(x) \Rightarrow \lim _{x \rightarrow 3} f(x)$ DNE

Let $F(x)=\frac{\sin x}{x}, \quad x \neq 0$

and $F(x) = 1. \quad x=0$

Check the Continuity at $x=0$

(1) $f(0)=1$

(2) $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

$f(0)=\lim _{x \rightarrow 0} f(x)$

at $x=0 \rightarrow f$ is Continuous

Show that the function $f(x)=2 \sqrt{3-x}$ is Continuous on the interval $(-\infty, 3]$

(1) $D f : 3-x \geq 0 \quad, \quad x \leq 3 \quad(-\infty, 3]$

(2) $\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a}[2 \sqrt{3-x}]=2 \sqrt{3-a}$

$f(a)=2 \sqrt{3-a}$

$\lim _{x \rightarrow a} f(x)=f(a)$

$f$ is Continuous $\forall \: a \in(-\infty, 3)$

(3) End Points

$x=3^{-} \quad \rightarrow \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} 2 \sqrt{3-x}=0$

$f(3)=2 \sqrt{3-3}=0$

$\lim _{x \rightarrow 3^{-}} f(x)=f(3)=0$

F is Contimuous at $x=3^{-}$

$f$ is Continuous on $(-\infty, 3]$