Evaluate the following one sided limits-i lim x - 2-x-3-x2-4x-3ii lim x - 2-x-3-x2-4iii lim x - 0-1-3 xiv lim x - 8-2 x-2-8vi lim x -2tan xvii lim x -2- xviii lim x - 0-x2-3 x-2-x3-2 x2ix lim x -2- x 2-1-2 x -4x lim x - 0-2- x xi lim x - 0- 1-cosecx

(i)lim_x → 2^+x 3/x^2 4 Put x =2+h ⇒ h =x 2 as x → 2^+ ⇒ x gt; 2 slightly ⇒ h gt; 0 ⇒ h → 0^+. =lim_h → 0^+(2+h) 3/(2+h)^2 2^2 =lim_h → 0^+2 3+h/(2+h 2)( ...

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Byju's Answer

Standard XII

Mathematics

Right Hand Limit

Evaluate the ...

Question

Evaluate the following one sided limits:

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

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

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

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

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

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

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

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

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

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

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

Open in App

Solution

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

Put x =2+h

$\Rightarrow h = x - 2$

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

$\Rightarrow h > 0$

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

$= {lim}_{h \to 0^{+}} \frac{(2 + h) - 3}{(2 + h)^{2} - 2^{2}}$

$= {lim}_{h \to 0^{+}} \frac{2 - 3 + h}{(2 + h - 2) (2 + h + 2)}$

$= {lim}_{h \to 0^{+}} \frac{(h - 1)}{(h) (4 + h)}$

$= {lim}_{h \to 0^{+}} \frac{1 - \frac{1}{h}}{4 + h}$

$= \frac{1 - \frac{1}{0}}{4} = - \infty .$

$(i i) {lim}_{x \to 2^{-}} \frac{x - 3}{x^{2} - 4} = {lim}_{x \to 0^{+}} \frac{(2 - h) - 3}{(2 - h)^{2} - 4}$

Put $x = 2 - h \Rightarrow h = 2 - x a s x \to 2^{-} \Rightarrow x < 2$ slightly

$\Rightarrow h > 0$

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

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

$= {lim}_{x \to 0^{+}} \frac{- 1 - h}{(4 - h) (- h)}$

$= {lim}_{x \to 0^{+}} \frac{\frac{1}{h} + 1}{(4 - h)}$

$= \frac{\frac{1}{0} + 4}{4} = \infty .$

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

Put $x = 0 + h \Rightarrow h = x$

as $x \to 0^{+} \Rightarrow 0^{+}$

$= {lim}_{h \to 0^{10}} \frac{1}{3 (0 + h)}$

$= {lim}_{h \to 0^{+}} \frac{1}{0 + 3 h}$

$= \frac{1}{0} = \infty$

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

Put $x = - 8 + h \Rightarrow h = x + 8$

as $\to - 8^{+} \Rightarrow x > - 8$

$\Rightarrow x + > 0$

$\Rightarrow h > 0^{+}$

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

$= {lim}_{h \to 0^{+}} \frac{2 (- 8 + h)}{(- 8 + h) + 8}$

$= {lim}_{h \to 0^{+}} \frac{- 16 + 2 h}{h}$

$= {lim}_{h \to 0^{+}} \frac{- 16}{h} + 2$

$\Rightarrow \frac{- 16}{0} + 2 = - \infty$

Put $x = 0 + h \Rightarrow h = x$

as $x \to 0^{+} \Rightarrow h \to 9^{+}$

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

$= \frac{2}{0} = \infty (v i) {lim}_{x \to \frac{π^{-}}{2}} t a n x$

Put $x = \frac{π}{2} - h \Rightarrow h = \frac{π}{2} - x$

as $x \to \frac{π^{-}}{2} \Rightarrow x <\to \frac{π}{2} s l i g h t l y$

$\Rightarrow \frac{π}{2} - x > 0$

$\Rightarrow h > 0$

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

$= {lim}_{h \to 0^{+}} t a n (\frac{π}{2} - h) = t a n (\frac{π}{2} - 0)$

$= t a n \frac{π}{2} = \infty$

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

Put $x = - \frac{π}{2} + h \Rightarrow h = x + \frac{π}{2}$

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

$\Rightarrow x + \frac{π}{2} > 0$

$\Rightarrow h > 0$

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

$= {lim}_{h \to 0^{+}} s e c (- \frac{π}{2} + h)$

$= s e c (- \frac{π}{2} + 0)$

$= \frac{1}{c o s (- \frac{π}{2})}$

$= \frac{- 1}{c o s (\frac{π}{2})}$

$= \frac{- 1}{0} = - \infty$

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

$= {lim}_{x \to 0^{-}} \frac{x^{2} - x - 2 x + 2}{x 62 (x - 2)}$

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

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

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

Put $x = 0 - h \Rightarrow h = - x$

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

$\Rightarrow - x > 0, h > 0 \Rightarrow h \to 0^{+}$

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

$= \frac{- h}{h^{2}} = \frac{- 1}{h} = \frac{- 1}{0} = - \infty$

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

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

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

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

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

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

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

$= {lim}_{h \to 0^{+}} \frac{(- 3 + h) (h - 1)}{2 h}$

$\Rightarrow {lim}_{h \to 0^{+}} \frac{- 3 x - 1}{2 \times 0} = \frac{1}{0} = \infty$

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

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

Put $x = 0 - h \Rightarrow h = - x$

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

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

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

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

$= {lim}_{h \to 0^{+}} 2 + \frac{1}{t a n h}$

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

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

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

Put $x = 0 - h \Rightarrow h = - x$

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

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

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

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

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

$= 1 - \frac{1}{0} = - \infty$

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Similar questions

Q. Evaluate the following one sided limits:

(i)

\lim_{x \to 2^{+}} \frac{x - 3}{x^{2} - 4}

(ii)

\lim_{x \to 2^{-}} \frac{x - 3}{x^{2} - 4}

(iii)

\lim_{x \to 0^{+}} \frac{1}{3 x}

(iv)

\lim_{x \to - 8^{+}} \frac{2 x}{x + 8}

(v)

\lim_{x \to 0^{+}} \frac{2}{x^{1 / 5}}

(vi)

\lim_{x \to \frac{π}{2}} \tan x

(vii)

\lim_{x \to - π / 2^{+}} \sec x

(viii)

\lim_{x \to 0^{-}} \frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}

(ix)

\lim_{x \to - 2^{+}} \frac{x^{2} - 1}{2 x + 4}

(x)

\lim_{x \to 0^{-}} 2 - \cot x

(xi)

\lim_{x \to 0^{-}} 1 + cosec x

Q. Find the limits :

i .

lim x \to 1 [\frac{x^{2} + 1}{x + 100}]

i i .

lim x \to 2 [\frac{x^{3} - 4 x^{2} + 4 x}{x^{2} - 4}]

i i i .

lim x \to 2 [\frac{x^{2} - 4}{x^{3} - 4 x^{2} + 4 x}]

i v .

lim x \to 2 [\frac{x^{3} - 2 x^{2}}{x^{2} - 5 x + 6}]

v .

lim x \to 1 [\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}]

Q. Evaluate the following:

\begin{matrix} {lim}_{x \to 1} \frac{x^{2} - 3 x + 2}{x^{2} - 4 x + 3} {lim}_{x \to 2} \frac{x^{3} + x^{2} - 12}{x^{3} - x^{2} - x - 2} {lim}_{x \to \sqrt{2}} \frac{x^{2} + x \sqrt{2} - 4}{x^{2} - 3 x \sqrt{2} + 4} {lim}_{x \to 3} \frac{x^{3} - 3 x^{2} - x + 3}{x^{3} - 6 x^{2} + 9 x} \end{matrix}

Q. STATEMENT-1 :

{lim}_{x \to \infty} (\frac{1^{2}}{x^{3}} + \frac{2^{2}}{x^{3}} + \frac{3^{2}}{x^{3}} + . . . . . . . + \frac{x^{2}}{x^{3}}) = lim x \to \infty \frac{1^{2}}{x^{3}} + lim x \to \infty \frac{2^{2}}{x^{3}} + . . . . . + lim x \to \infty \frac{x^{2}}{x^{3}} = 0

STATEMENT-2 :

{lim}_{x \to a} (f_{1} (x) + f_{2} (x) + . . . . . + f_{n} (x)) = {lim}_{x \to a} f_{1} (x) + {lim}_{x \to a} f_{2} (x) + . . . . + {lim}_{x \to a} f_{n} (x)

, where

n \in N .

${lim}_{x \to 1} {\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}}$

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