Lim x - 1x-2-x2-x-1-x3-3 x2-2 x

Lim_x → 1{x 2/x^2 x 1/x^3 3x^2+2x} =lim_x → 1{x 2/x(x 1) 1/x(x^2 3x+2)} =lim_x → 1{x 2/x(x 1) 1/x(x^2 1x 2x+2)} =lim_x → 1{x 2/x(x 1) 1/x(x^2 1x 2x+2)} =lim_x → ...

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Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

lim x → 1x-2/...

Question

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

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Solution

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

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

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

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

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

$= {lim}_{x \to 1} \frac{1}{x (x - 1)} [\frac{x - 2}{1} - \frac{1}{x - 2}]$

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

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

$[∵ a^{2} - b^{2} = (a - b) (a + b)]$

$= {lim}_{x \to 1} \frac{(x - 3) (x - 1)}{x (x - 1) (x - 2)}$ [It is of $(\frac{0}{0})$ form]

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

$= \frac{(1 - 3)}{1 (1 - 2)} = \frac{- 2}{- 1} = 2$

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Similar questions

Solve:

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

Q. Find the limits :

i .

lim x \to 1 [\frac{x^{2} + 1}{x + 100}]

i i .

lim x \to 2 [\frac{x^{3} - 4 x^{2} + 4 x}{x^{2} - 4}]

i i i .

lim x \to 2 [\frac{x^{2} - 4}{x^{3} - 4 x^{2} + 4 x}]

i v .

lim x \to 2 [\frac{x^{3} - 2 x^{2}}{x^{2} - 5 x + 6}]

v .

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

Q. The value of

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

lim x \to 2 [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x^{3} - 3 x^{2} + 2 x}]

Q. Which of the following are quadratic equations?

(i) x² + 6x − 4 = 0
(ii)

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

(iii)

x^{2} + \frac{1}{x^{2}} = 5

(iv)

x - \frac{3}{x} = x^{2}

(v)

2 x^{2} - \sqrt{3 x} + 9 = 0

(vi)

x^{2} - 2 x - \sqrt{x} - 5 = 0

(vii) 3x² − 5x + 9 = x² − 7x + 3
(viii)

x + \frac{1}{x} = 1

(ix) x² − 3x = 0
(x)

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)

x + \frac{1}{x} = x^{2}, x \neq 0

(xiii) 16x² − 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)³ = x³ − 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)

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limx→1{x−2x2−x−1x3−3x2+2x}

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