Lim_x → 1x 1/√(x^2+3) 2 Rationalising the denominator =lim_x → 1(x 1)× (√(x^2+3)+2)/(√(x^2+3) 2)(√(x^2+3)+2) =lim_x → 1(x 1)× (√(x^2+3)+2)/(x^2+3 4) =lim_x → 1( ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x → 1x-1/...

Question

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

Open in App

Solution

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

Rationalising the denominator

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

$= {lim}_{x \to 1} \frac{(x - 1) \times (\sqrt{x^{2} + 3} + 2)}{(x^{2} + 3 - 4)}$

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

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

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

$\Rightarrow \frac{\sqrt{1 + 3} + 2}{1 + 1}$

$= \frac{2 + 2}{2}$

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

Suggest Corrections

Similar questions

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

Q. Evaluate:
1.

{lim}_{x \to 2} \frac{x^{3} - 3 x^{2} + 4}{x^{4} - 8 x^{2} + 16}

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

{lim}_{x \to - 3} [\frac{1}{x^{2} + 4 x + 3} + \frac{1}{x^{2} + 8 x + 15}]

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

lim x \to 1 (\frac{x + 2}{x^{2}} + 1)

is equal to?

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

Join BYJU'S Learning Program

limx→1x−1√x2+3−2

limx→1x−1√x2+3−2 Rationalising the denominator =limx→1(x−1)×(√x2+3+2)(√x2+3−2)(√x2+3+2) =limx→1(x−1)×(√x2+3+2)(x2+3−4) =limx→1(x−1)×(√x2+3+2)(x2−1) =limx→1(x−1)×(√x2+3+2)(x−1)(x+1) =limx→1√x2+3+2x+1 ⇒√1+3+21+1 =2+22 =42=2

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