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

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x → 2√x2+...

Question

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

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Solution

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

Rationalising the numenator

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

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

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

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

$= \frac{(2 + 2)}{\sqrt{4 + 1} + \sqrt{5}}$

$= \frac{4}{2 \sqrt{5}} = \frac{2}{\sqrt{5}}$

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Similar questions

Evaluate the following one sided limits:

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

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

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

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

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

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

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

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

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

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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^{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}]

lim x \to \infty \frac{2 x^{2} + x + 5}{x^{2} + x - 2}

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

lim x \to - 1 \frac{x + 1}{2 - \sqrt{4 + x + x^{2}}} =

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limx→2√x2+1−√5x−2

limx→2√x2+1−√5x−2 Rationalising the numenator =limx→2√x2+1−√5(x−2)×(√x2+1+√5)(√x2+1+√5) =limx→2(x2+1−5)(x−2)(√x2+1+√5) =limx→2(x+2)(x−2)(x−2)(√x2+1+√5) =limx→2(x+2)(√x2+1+√5) =(2+2)√4+1+√5 =42√5=2√5

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