Lim x - 11 - 2-x-2-x x 3-1

nbsp;limx #8594;1x3 1 xx2+x 2x2+x 2x3 1=limx #8594;1x3 1 x3 x2+2xx2+x 2x3 1=limx #8594;1 x2+2x 1x2+x 2x 1x2+x+1=limx #8594;1 x2 2x+1x2+x 2x 1x2+x+1=l ...

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Byju's Answer

Standard XII

Mathematics

Property 4

lim x → 11 × ...

Question

$\lim_{x \to 1} (\frac{1}{x^{2} + x - 2} - \frac{x}{x^{3} - 1})$

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Solution

$\lim_{x \to 1} [\frac{(x^{3} - 1) - x (x^{2} + x - 2)}{(x^{2} + x - 2) (x^{3} - 1)}] = \lim_{x \to 1} [\frac{(x^{3} - 1) - x^{3} - x^{2} + 2 x}{(x^{2} + x - 2) (x^{3} - 1)}] = \lim_{x \to 1} [\frac{- x^{2} + 2 x - 1}{(x^{2} + x - 2) (x - 1) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x^{2} - 2 x + 1)}{(x^{2} + x - 2) (x - 1) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- {(x - 1)}^{2}}{(x^{2} + x - 2) (x - 1) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x - 1)}{(x^{2} + 2 x - x - 2) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x - 1)}{\{x (x + 2) - 1 (x + 2)\} (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x - 1)}{(x - 1) (x + 2) (x^{2} + x + 1)}] = \frac{- 1}{(1 + 2) (1 + 1 + 1)} = \frac{- 1}{9}$

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limx→1 1x2+x-2-xx3-1

limx→1x3-1-xx2+x-2x2+x-2x3-1=limx→1x3-1-x3-x2+2xx2+x-2x3-1=limx→1-x2+2x-1x2+x-2x-1x2+x+1=limx→1-x2-2x+1x2+x-2x-1x2+x+1=limx→1-x-12x2+x-2x-1x2+x+1=limx→1-x-1x2+2x-x-2x2+x+1=limx→1-x-1xx+2-1x+2x2+x+1=limx→1-x-1x-1x+2x2+x+1=-11+21+1+1=-19

$\lim_{x \to 1} (\frac{1}{x^{2} + x - 2} - \frac{x}{x^{3} - 1})$