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| ...

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Byju's Answer

Standard XI

Mathematics

Relation between Continuity and Differentiability

0, & x =0lend...

Question

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

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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.

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