Question

limit x tends to 1
[(x-2/x^2-x)-1/x^3-3x^2+2x)]

Answer 1

$\textbf{Given:}$

$\displaystyle\lim_{x\to\;1}[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}]$

$\textbf{To find:}$

$\text{The value of \displaystyle\lim_{x\to\;1}[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}] }$

$\textbf{Solution:}$

$\text{Consider,}$

$\displaystyle\lim_{x\to\;1}[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}]$

$=\displaystyle\lim_{x\to\;1}[\dfrac{x-2}{x(x-1)}-\dfrac{1}{x(x^2-3x+2)}]$

$=\displaystyle\lim_{x\to\;1}[\dfrac{x-2}{x(x-1)}-\dfrac{1}{x(x-1)(x-2)}]$

$=\displaystyle\lim_{x\to\;1}\dfrac{1}{x(x-1)}[x-2-\dfrac{1}{x-2}]$

$=\displaystyle\lim_{x\to\;1}\dfrac{1}{x(x-1)}[\dfrac{(x-2)^2-1}{x-2}]$

$=\displaystyle\lim_{x\to\;1}\dfrac{1}{x(x-1)}[\dfrac{x^2-4x+4-1}{x-2}]$

$=\displaystyle\lim_{x\to\;1}\dfrac{1}{x(x-1)}[\dfrac{x^2-4x+3}{x-2}]$

$=\displaystyle\lim_{x\to\;1}\dfrac{1}{x(x-1)}[\dfrac{(x-1)(x-3)}{x-2}]$

$=\displaystyle\lim_{x\to\;1}\dfrac{1}{x}[\dfrac{x-3}{x-2}]$

$=\displaystyle\lim_{x\to\;1}\;\dfrac{x-3}{x(x-2)}$

$=\displaystyle\dfrac{\lim_{x\to\;1}\;(x-3)}{\lim_{x\to\;1}\;x(x-2)}$

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

$=\dfrac{-2}{-1}$

$=2$

$\therefore\bf\,\displaystyle\lim_{x\to\;1}[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}]=2$

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Answer 2

Answer:

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