Question

$\huge\red{\boxed{\sf QuEsTioN}}$


limx→ 0 |x|/x is equal to:

(a)1 (b)-1 (c)0 (d)does not exists​

Answer 1

Question :-

$\displaystyle\lim_{x \to 0 } \rm{\frac{ |x| }{x} }$

Solution :-

$\longrightarrow\rm{for \: x < 0, \frac{ |x| }{x} = \frac{ - x}{x} = - 1 } \\ \\ \rm \longrightarrow{for \: x > 0, \frac{ |x| }{x} = \frac{x}{x} = 1 } \: \: \: \: \: \:$

Thus,

$\longrightarrow \rm{ \displaystyle \lim_{x \to0 {}^{ - } } \rm{\frac{ |x| }{x} = - 1 }} \\ \\ \longrightarrow \rm{ \displaystyle \lim_{x \to0 {}^{ + } } \rm{\frac{ |x| }{x} = 1 }} \: \: \:$

So limit doesn't exist.

Option (D) is the right answer.

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Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ |x| }{x}$

We know,

Limit of a function f(x) at x = a exist iff

$\boxed{ \rm \:\displaystyle\lim_{x \to a^ + } \: f(x) \: = \: \displaystyle\lim_{x \to a^ - } \: f(x)}$

Now, Consider

Left Hand Limit

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0^ - } \frac{ |x| }{x}$

To evaluate this,

$\red{\rm :\longmapsto\:Put \: x = 0 - h = - h}$

$\red{\rm :\longmapsto\:As \: x \: \to \: 0 \: \: so \: h \: \to \: 0}$

So, above limit can be rewritten as

$\rm \: = \: \displaystyle\lim_{h \to 0} \: \frac{ |0 - h| }{ - h}$

$\rm \: = \: \displaystyle\lim_{h \to 0} \: \frac{ |- h| }{ - h}$

$\rm \: = \: \displaystyle\lim_{h \to 0} \: \frac{h}{ - h}$

$\rm \: = \: - \: 1$

So,

$\rm :\longmapsto\:\boxed{ \rm \:\displaystyle\lim_{x \to 0^ - } \frac{ |x| }{x} \: = \: - \: 1} - - - (1)$

Now, Consider

Right Hand Limit

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0^ + } \frac{ |x| }{x}$

To evaluate this limit,

$\red{\rm :\longmapsto\:Put \: x = 0 + h = h}$

$\red{\rm :\longmapsto\:As \: x \: \to \: 0 \: \: so \: h \: \to \: 0}$

So, above limit can be rewritten as

$\rm \: = \: \displaystyle\lim_{h \to 0} \: \frac{ |0 + h| }{0 + h}$

$\rm \: = \: \displaystyle\lim_{h \to 0} \: \frac{ |h| }{ h}$

$\rm \: = \: \displaystyle\lim_{h \to 0} \: \frac{ h}{ h}$

$\rm \: = \: 1$

So,

$\rm :\longmapsto\:\boxed{ \rm \:\displaystyle\lim_{x \to 0^ + } \frac{ |x| }{x} \: = \: 1} - - - (2)$

So, from equation (1) and (2), we concluded that

$\rm :\longmapsto\:\displaystyle\lim_{h \to 0^ - } \frac{ |x| }{x} \: \ne \: \displaystyle\lim_{h \to 0^ + } \frac{ |x| }{x}$

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ |x| }{x} \sf \: does \: not \: exist.$

• So, Option (d) is correct