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