Lim x - 1-3-x-5-x-x2-1

Lim_x → 1√(3+x) √(5 x)/x^2 1 Rationalising the numenator =lim_x → 1(√(3+x) √(5 x))/(x^2 1)×(√(3+x) √(5 x))/(√(3+x)+√(5 x)) =lim_x → 1(3+x) (5 x)/(x 1)(x+1)(√(3+ ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x → 1√3+x...

Question

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

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Solution

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

Rationalising the numenator

$= {lim}_{x \to 1} \frac{(\sqrt{3 + x} - \sqrt{5 - x})}{(x^{2} - 1)} \times \frac{(\sqrt{3 + x} - \sqrt{5 - x})}{(\sqrt{3 + x} + \sqrt{5 - x})}$

$= {lim}_{x \to 1} \frac{(3 + x) - (5 - x)}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}$

$= {lim}_{x \to 1} \frac{- 2 + 2 x}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}$

$= {lim}_{x \to 1} \frac{2}{(x + 1) (\sqrt{3} = x + \sqrt{5 - x})}$

$= \frac{2}{(1 + 1) (\sqrt{3 + 1} + \sqrt{5 - 1})} = \frac{2}{(2) (2 + 2)}$

$= \frac{1}{4}$

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Similar questions

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

Q. Evaluate

lim x \to 1 \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}

Show that ${lim}_{x \to \infty} (\sqrt{x^{2} + x + 1 - x}) \neq {lim}_{x \to \infty} (\sqrt{x^{2} + 1} - x)$

Q. Show that

\lim_{x \to \infty} (\sqrt{x^{2} + x + 1} - x) \neq \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x)

Evaluate the following one sided limits:

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

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

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

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

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

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

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

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

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

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

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

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limx→1√3+x−√5−xx2−1

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