Lim x - 2-1-4 x-5 x-2 x-x-2

Lim_x → 2√(1+4x) √(5x+2x)/x 2 Rationalising the numenator lim_x → 2(√(1+4x) √(5+2x))/(x 2)×(√(1+4x)+√(5+2x))/(√(1+4x)+√(5+2x)) =lim_x → 2(1+4x) (5+2x)/(x 2)(√(1 ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x → 2√1+4...

Question

${lim}_{x \to 2} \frac{\sqrt{1 + 4 x} - \sqrt{5 x + 2 x}}{x - 2}$

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Solution

${lim}_{x \to 2} \frac{\sqrt{1 + 4 x} - \sqrt{5 x + 2 x}}{x - 2}$

Rationalising the numenator

${lim}_{x \to 2} \frac{(\sqrt{1 + 4 x} - \sqrt{5 + 2 x})}{(x - 2)} \times \frac{(\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}{(\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}$

$= {lim}_{x \to 2} \frac{(1 + 4 x) - (5 + 2 x)}{(x - 2) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}$

$= {lim}_{x \to 2} \frac{- 4 + 2 x}{(x - 2) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}$

$= {lim}_{x \to 2} \frac{2 (x - 2)}{(x - 2) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}$

$= \frac{2}{\sqrt{1 + 8} + \sqrt{5 + 4}} = \frac{2}{\sqrt{9} + \sqrt{9}}$

$= \frac{2}{6} = \frac{1}{3}$

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Similar questions

Q. Find the limits :

i .

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

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

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

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Q. Evaluate:

lim x \to 2 \frac{\sqrt{5 x + 2} - \sqrt{4 x + 4}}{x - 2}

${lim}_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}$

Evaluate the following one sided limits:

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

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

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

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

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

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

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

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

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

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

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

lim x \to 2 ⎛ ⎝ {(\frac{x^{3} - 4 x}{x^{3} - 8})}^{- 1} - {(\frac{x + \sqrt{2 x}}{x - 2} - \frac{\sqrt{2}}{\sqrt{x} - \sqrt{2}})}^{- 1} ⎞ ⎠

is equal to

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limx→2√1+4x−√5x+2xx−2

limx→2√1+4x−√5x+2xx−2 Rationalising the numenator limx→2(√1+4x−√5+2x)(x−2)×(√1+4x+√5+2x)(√1+4x+√5+2x) =limx→2(1+4x)−(5+2x)(x−2)(√1+4x+√5+2x) =limx→2−4+2x(x−2)(√1+4x+√5+2x) =limx→22(x−2)(x−2)(√1+4x+√5+2x) =2√1+8+√5+4=2√9+√9 =26=13

${lim}_{x \to 2} \frac{\sqrt{1 + 4 x} - \sqrt{5 x + 2 x}}{x - 2}$