Question

show that f(x) =|x-1| is countinous at x=1​

Answer 1

Given Question :-

• Show that f(x) =|x-1| is countinous at x=1

ANSWER

Let us first define a function,

• f(x) =|x-1|

$\begin{gathered}\begin{gathered}\bf \: |x - 1| = \begin{cases} &\sf{ - (x - 1) \: \: if \: x \: < \: 1} \\ &\sf{x - 1 \: \: if \: x \: \geqslant \: 1} \end{cases}\end{gathered}\end{gathered}$

$:\implies\:\begin{gathered}\begin{gathered}\bf \: |x - 1| = \begin{cases} &\sf{ 1 - x \: \: if \: x \: < \: 1} \\ &\sf{x - 1 \: \: if \: x \: \geqslant \: 1} \end{cases}\end{gathered}\end{gathered}$

Now,

• we have to check the continuity of the f(x) at x = 1.

Now,

$\longmapsto \rm \: f(1) \: = \: 1 - 1$

$\rm :\implies\:f(1) \: = \: 0 - - (1)$

Now,

• Consider LHL,

$\longmapsto \rm \: \tt \:\lim_{x\to1 ^{ - } } \: f(x)$

$\longmapsto \rm \: \tt \:\lim_{x\to1 ^{ - } }(1 - x)$

$\pink{ \tt \: Put \: x = 1 - h, as \: x \to \: 1 \: so \: h \to \: 0}$

$\longmapsto \rm \: \tt \:\lim_{h\to0}(1 - 1 + h)$

$\longmapsto \rm \: \tt \:\lim_{h\to0}(h)$

$\longmapsto \rm \: 0 - - (2)$

Now,

• Consider RHL,

$\longmapsto \rm \: \tt \:\lim_{x\to1 ^{ + } } \: f(x)$

$\longmapsto \rm \: \tt \:\lim_{x\to1 ^{ + } } \: (x - 1)$

$\pink{ \tt \: Put \: x = 1 + h, as \: x \to \: 1 \: so \: h \to \: 0}$

$\longmapsto \rm \: \tt \:\lim_{h\to0}(1 + h - 1)$

$\longmapsto \rm \: \tt \:\lim_{h\to0}(h)$

$\longmapsto \rm \: 0 - - (3)$

So,

From (1), (2) and (3), we concluded that

$\longmapsto \rm \: \tt \:\lim_{x\to1 ^{ - } } = \:\lim_{x\to1 ^{ - } } \: = \: f(1)$

$\boxed{ \implies \green{ \bf \:Hence, \: f(x) \: = \: |x-1| \: is \: countinous \: at \: x \: = \: 1 }}$