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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 $$(-\infty, 3]$$

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