Show that lim x - 2-x-x-lim x - 2-x-x-

Lim_x → 2^ x/[x] Let x =2 h h =2 x as x → 2 ⇒ x lt; 2 slightly ⇒ 2 x gt; 0 ⇒ h gt; 0 ⇒ 4 → 0^+ =lim_h → 0^+2 h/[2 h] =lim_h → 0^+2 h/1=lim_h → 0^+(2 ...

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Standard XII

Mathematics

Existence of Limit

Show that lim...

Question

Show that ${lim}_{x \to 2^{-}} \frac{x}{[x]} \neq {lim}_{x \to 2^{+}} \frac{x}{[x]}$ .

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Solution

${lim}_{x \to 2^{-}} \frac{x}{[x]}$

Let x =2-h

h =2-x

as $x \to 2 - \Rightarrow x < 2$ slightly

$\Rightarrow 2 - x > 0$

$\Rightarrow h > 0 \Rightarrow 4 \to 0^{+}$

$= {lim}_{h \to 0^{+}} \frac{2 - h}{[2 - h]}$

$= {lim}_{h \to 0^{+}} \frac{2 - h}{1} = {lim}_{h \to 0^{+}} (2 - h) = 2$

Now ${lim}_{x \to 2^{+}} \frac{x}{(x)}$

Let $x = 2 + h \Rightarrow h = x - 2$

as $x \to 2^{+} \Rightarrow x > 2$ slightly

$\Rightarrow x - 2 > 0$

$\Rightarrow h > 0^{+}$

$= {lim}_{h \to 0^{+}} \frac{2 + h}{[2 + h]} = {lim}_{h \to 0^{+}} \frac{2 + h}{2}$

$= \frac{2 + 0}{2} = 1$

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Show that limx→2−x[x]≠limx→2+x[x].

limx→2−x[x] Let x =2-h h =2-x as x→2−⇒x<2 slightly ⇒2−x>0 ⇒h>0⇒4→0+ =limh→0+2−h[2−h] =limh→0+2−h1=limh→0+(2−h)=2 Now limx→2+x(x) Let x=2+h⇒h=x−2 as x→2+⇒x>2 slightly ⇒x−2>0 ⇒h>0+ =limh→0+2+h[2+h]=limh→0+2+h2 =2+02=1

Show that ${lim}_{x \to 2^{-}} \frac{x}{[x]} \neq {lim}_{x \to 2^{+}} \frac{x}{[x]}$ .