Lim x - 12-1-x2-1-x-1-

The correct option is A 1/2 lim_x → 1(2/1 x^2+1/x 1) = lim_x → 1(2/1 x^2 1/1 x) = lim_x → 1(2 (1+x)/1 x^2) =lim_x → 1(1 x/1 x^2) = lim_x → 11/1+x = 1/2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Limit

lim x → 12/1-...

Question

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{1}{2}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\infty$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$\frac{1}{2}$

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

Suggest Corrections

Similar questions

lim x \to - 1 {(\frac{x^{4} + x^{2} + x + 1}{x^{2} - x + 1})}^{\frac{1 - c o s (x + 1)}{(x + 1)^{2}}}

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

lim x \to - 1 {(\frac{x^{4} + x^{2} + x + 1}{x^{2} - x + 1})}^{\frac{1 - cos (x + 1)}{(x + 1)^{2}}}

is equal to

lim x \to 1 {(\frac{x^{3} + 2 x^{2} + x + 1}{x^{2} + 2 x + 3})}^{\frac{1 - cos (x - 1)}{(x - 1)^{2}}} =

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

Join BYJU'S Learning Program

limx→1(21−x2+1x−1)=___

The correct option is B 12 limx→1(21−x2+1x−1)=limx→1(21−x2−11−x)=limx→1(2−(1+x)1−x2)=limx→1(1−x1−x2)=limx→111+x=12

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

The correct option is B
$\frac{1}{2}$

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