0- - x -0lendmatrix right-

Here x→ 0lim = { |x/x|, amp; x ≠ 0 0, amp; x = 0 . L.H.L. = x → 0^ lim f(x) = x → 0^ limx|x| Put x=0 h as x → 0, h → 0 ∴h → 0lim|0 h|/0 h = h → 0lim| ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Relation between Continuity and Differentiability

0, & x =0lend...

Question

Evaluate $lim x \to 0 = {\begin{matrix} | \frac{x}{x} |, & x \neq 0 0, & x = 0 \end{matrix}$

Open in App

Solution

Here $lim x \to 0 = {\begin{matrix} | \frac{x}{x} |, & x \neq 0 0, & x = 0 \end{matrix}$

L.H.L. = $lim x \to 0^{-} f (x) = lim x \to 0^{-} | x - - | x$

Put x=0 - h as $x \to 0, h \to 0$

$∴ lim h \to 0 \frac{| 0 - h |}{0 - h} = lim h \to 0 \frac{| - h |}{- h} = - 1$

R.H.L. = $lim x \to 0^{+} f (x) = lim x \to 0^{+} \frac{h}{h} = 1$

Now L.H.L. $\neq$ R.H.L.

Thus limit does not exist at x =0.

Suggest Corrections

Similar questions

Find $lim x \to 0$ f(x) and $lim x \to 1$ f(x) where f(x)= ${\begin{matrix} 2 x + 3 & x \leq 0 3 (x + 1) & x > 0 \end{matrix}$

f(x)= ${\begin{matrix} \frac{| x |}{x}, i f x \neq 0 0, i f x = 0 \end{matrix}$

$f (x) = {\begin{matrix} x^{2} s i n \frac{1}{x}, & i f x \neq 0 0, & i f x = 0 \end{matrix}$

Q. Let

f (x) = {\begin{matrix} x^{p} s i n \frac{1}{x}, & x \neq 0 0, & x = 0 \end{matrix}

then f(x) is continuous but not differential at x = 0 if

Q. If

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x - 1 & x < 0 \frac{1}{4}, & x = 0 x^{2}, & x > 0 \end{matrix}

, then

Join BYJU'S Learning Program

Evaluate limx→0={|xx|,x≠00,x=0

Here limx→0={|xx|,x≠00,x=0 L.H.L. = limx→0−f(x)=limx→0−|x––|x Put x=0 - h as x→0,h→0 ∴limh→0|0−h|0−h=limh→0|−h|−h=−1 R.H.L. = limx→0+f(x)=limx→0+hh=1 Now L.H.L. ≠ R.H.L. Thus limit does not exist at x =0.

Evaluate $lim x \to 0 = {\begin{matrix} | \frac{x}{x} |, & x \neq 0 0, & x = 0 \end{matrix}$