Is the function f defined by f(x) = x, if x 1 5, if x > 1 continuous at x = 0? At x = 1? At x = 2?

Here f(x) = x, if x 1 5, if x > 1 At x = 0 ^ Lt _ x 0^ - f(x) = ^ Lt _ x 0 (x) [ Put x = 0 - h, h>0 so that h 0 as x 0^-] = ^ Lt _ h 0 (0-h) = ^ Lt _ h 0 (-h) = (0) = 0 ^ Lt _ x 0^ + f(x) = ^ Lt _ x 0^ + (x) [ Put x = 0 + h, h>0 so that h 0 as x 0^+] = ^ Lt _ h 0 (0 + h) = 0 + 0 = 0 ^ Lt _ x 0^ - f(x) = ^ Lt _ x 0^ + f(x) = 0 ^ Lt _ x 0 f(x) = 0 Also f(0) = value of x at x = 0 = 0 f is continous at x = 0. At x = 1 ^ Lt _ x 1^ - f(x) = ^ Lt _ x 1^ - x [ Put x = 1 - h, h>0, so that h 0 on x 1^-] ^ Lt _ h 0 (1 - h) = 1 - 0 = 1 ^ Lt _ x 1^ + f(x) = ^ Lt _ x 1^ + (5) = 5 ^ Lt _ x 1^ - f(x) ^ Lt _ x 1^ + f(x) ^ Lt _ x 1^ - f(x) does not exist f is discontinuous at x =1. At x = 2 ^ Lt _ x 2 f(x) = ^ Lt _ x 2 (5) = 5 Also f(2) = 5 ^ Lt _ x 2 f(x) = f(2) = 5 f is continous at x = 2.

Is the function f defined by f(x) = x, if x 1 5, if x > 1 continu…

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Question

Is the function f defined by
$\text{f(x)} = \begin{cases}\text{x}, \text{if}\ \text{x}\leq1\\5, \text{if}\ \text{x} > 1\end{cases}$
continuous at x = 0? At x = 1? At x = 2?

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Answer

Here $\text{f(x)} = \begin{cases}\text{x}, \text{if}\ \text{x}\leq1\\5, \text{if}\ \text{x} > 1\end{cases}$At x = 0
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\text{(x)}\ [ \text{Put}\ \text{x} = 0 - \text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow 0\ \text{as}\ \text{x}\rightarrow 0^-]$
$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(0-\text{h})$
$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(-h)} = (0) = 0$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{(x)}\ [ \text{Put}\ \text{x} = 0 + \text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow 0\ \text{as}\ \text{x}\rightarrow 0^+]$
$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(0 + \text{h})$
= 0 + 0 = 0
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = 0$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\text{f(x)} = 0$
Also f(0) = value of x at x = 0
= 0
$\therefore$ f is continous at x = 0.
At x = 1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{x}\ [ \text{Put}\ \text{x} = 1 - \text{h}, \text{h}>0,\ \text{so that}\ \text{h}\rightarrow 0\ \text{on}\ \text{x}\rightarrow 1^-]$
$^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(1 - \text{h}) = 1 - 0 = 1$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}(5) = 5$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)}\neq\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)}$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)}$ does not exist
$\therefore$ f is discontinuous at x =1.
At x = 2
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}(5) = 5$
Also f(2) = 5
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}\text{f(x)} = \text{f(2)} = 5$
$\therefore$ f is continous at x = 2.

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