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 ...

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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}

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Solution

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

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