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

The correct option is D 1/2lim_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 ...

187

You visited us 187 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Expansions to Remove Indeterminate Form

lim x → 12/1-...

Question

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

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

Suggest Corrections

Similar questions

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

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}}} =

Join BYJU'S Learning Program

limx→1(21−x2+1x−1)=___

The correct option is A 12limx→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 A $\frac{1}{2}$
${lim}_{x \to 1} (\frac{2}{1 - x^{2}} + \frac{1}{x - 1}) = {lim}_{x \to 1} (\frac{2}{1 - x^{2}} - \frac{1}{1 - x}) = {lim}_{x \to 1} (\frac{2 - (1 + x)}{1 - x^{2}}) = {lim}_{x \to 1} (\frac{1 - x}{1 - x^{2}}) = {lim}_{x \to 1} \frac{1}{1 + x} = \frac{1}{2}$