Find-i lim x - 2- x -ii lim x - 5- x -iii lim x - 1- x -

(i)lim_x → 2 [x] LHL: lim_x → 2^ [X] Let x =2 h, ⇒ h=2 x as x → 2^0 ⇒ x lt; 2 slightly ⇒ 2 x gt;0 ⇒ h gt;0 ⇒ h → 0^+ lim_h → 0^+[2 h]=1 RHL lim_x → 2^+ ...

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Byju's Answer

Standard XII

Mathematics

Limit

Find:i lim x ...

Question

Find:

$(i) {lim}_{x \to 2} [x]$

$(i i) {lim}_{x \to \frac{5}{2}} [x]$

$(i i i) {lim}_{x \to 1} [x]$

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Solution

$(i) {lim}_{x \to 2} [x]$

LHL:

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

Let x =2-h,

$\Rightarrow h = 2 - x$

as $x \to 2^{0} \Rightarrow x < 2$ slightly

$\Rightarrow 2 - x > 0 \Rightarrow h > 0$

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

${lim}_{h \to 0^{+}} [2 - h] = 1$

$R H L {lim}_{x \to 2^{+}} [X]$

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

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

$\Rightarrow - 2 > 0 \Rightarrow h > 0$

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

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

$L H L \neq R H L$

Thus, ${lim}_{x \to 2} [x]$ does not exist.

$(i i) {lim}_{x \to \frac{5}{2}} [x]$

$R H L : {lim}_{x \to \frac{5^{+}}{2}} [x]$

Let $x = \frac{5}{2} + h,$

$\Rightarrow h = x - \frac{5}{2}$

as $x \to \frac{5^{+}}{2} \Rightarrow x > \frac{5}{2}$ slightly

$\Rightarrow x - \frac{5}{2} > 0$

$\Rightarrow h > 0$

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

${lim}_{h \to 0^{+}} [\frac{5}{2} + h] = 2$

Hence, ${lim}_{x \to \frac{5^{-}}{2}} [x]$

Let $x = \frac{5}{2} - h, \Rightarrow h = \frac{5}{2} - x$

as $x \to \frac{5^{-}}{2} \Rightarrow x < \frac{5}{2}$ slightly

$\Rightarrow \frac{5}{2} - x > 0$

$\Rightarrow h > 0$

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

${lim}_{h \to 0^{+}} [\frac{5}{2} - h] = 2$

Hence, ${lim}_{x \to \frac{5^{-}}{2}} [x] = 2$

Since, ${lim}_{x \to \frac{5^{-}}{2}} [x] = {lim}_{x \to \frac{5^{+}}{2}} [x]$

$\Rightarrow {lim}_{x \to \frac{5}{2}} [x] = 2$

$(i i i) {lim}_{x \to 1} [x]$

$L H L : {lim}_{x \to 1^{-}} [x]$

Let x =1-h,

$\Rightarrow h = 1 - x$

as $x \to 1^{-} \Rightarrow x < 1$ slightly

$\Rightarrow 1 - x > 0 \Rightarrow h > 0$

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

${lim}_{h \to 0^{+}} [1 - h]$

${lim}_{x \to 1^{-}} [x] = 0$

$R H L : {lim}_{x \to 1^{+}} [x]$

Let x =1+h,

$\Rightarrow h = x - 1$

as $x \to 1^{+} \Rightarrow x > 1$ slightly

$\Rightarrow x - 1 > 0 \Rightarrow h > 0$

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

${lim}_{x \to 0^{+}} [1 + 4] = 1$

$\Rightarrow {lim}_{x \to 1} [x] \neq {lim}_{x \to 1^{+}} [x]$

Thus, ${lim}_{x \to 1} [x]$ does not exist.

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Similar questions

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Q. 1)

lim x \to 0 \frac{[x]}{x}

lim x \to 1 \frac{[x]}{x}

Evaluate the following one sided limits:

$(i) {lim}_{x \to 2^{+}} \frac{x - 3}{x^{2} - 4}$

$(i i) {lim}_{x \to 2^{-}} \frac{x - 3}{x^{2} - 4}$

$(i i i) {lim}_{x \to 0^{+}} \frac{1}{3 x}$

$(i v) {lim}_{x \to 8^{+}} \frac{2 x}{x + 8}$

$(v) {lim}_{x \to 0^{+}} \frac{2}{x^{\frac{1}{5}}}$

$(v i) {lim}_{x \to \frac{π^{-}}{2}} t a n x$

$(v i i) {lim}_{x \to \frac{π}{2^{+}}} s e c x$

$(v i i i) {lim}_{x \to 0^{-}} \frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}$

$(i x) {lim}_{x \to - 2^{+}} \frac{x^{2} - 1}{2 x + 4}$

$(x) {lim}_{x \to 0^{+}} (2 - c o t x)$

$(x i) {lim}_{x \to 0^{-}} 1 + c o s e c x$

Q. Let

f (x) = [\frac{x}{[x]}], g (x) = [x [\frac{1}{x}]]

and

h (x) = [\frac{[x]}{x}],

, then

lim x \to 2^{-} f (x) + lim x \to {\frac{1}{2}}^{+} g (x) + lim x \to 2^{+} h (x) =

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Find: (i)limx→2[x] (ii)limx→52[x] (iii)limx→1[x]

Find:

$(i) {lim}_{x \to 2} [x]$

$(i i) {lim}_{x \to \frac{5}{2}} [x]$

$(i i i) {lim}_{x \to 1} [x]$