Question

evaluate lim x tend to 1 [(x-2/x^2-x) - 1/x^3-3x^2+2x]

Answer 1

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{x-2}{x^{2} -x}-\dfrac{1}{x^{3} -3x^{2} +2x} \bigg]$

The function can also be written as:

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{x-2}{x(x-1)}-\dfrac{1}{x(x^{2} -3x+2)} \bigg]$

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{x-2}{x(x-1)}-\dfrac{1}{x(x^{2} -2x-x+2)} \bigg]$

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{x-2}{x(x-1)}-\dfrac{1}{x(x -2)- 1(x-2)} \bigg]$

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{x-2}{x(x-1)}-\dfrac{1}{x(x -1)(x-2)} \bigg]$

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{(x-2)^2-1}{x(x -1)(x-2)} \bigg]$

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{(x^{2} -4x+4-1}{x(x -1)(x-2)} \bigg]$

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{(x^{2} -4x+3}{x(x -1)(x-2)} \bigg]$

Now, evaluate the above limit

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{(x^{2} -3x-x+3}{x(x -1)(x-2)} \bigg]$

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{(x(x -3)-1(x-3)}{x(x -1)(x-2)} \bigg]$

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{(x -1)(x-3)}{x(x -1)(x-2)} \bigg]$

$\\\quad\longrightarrow\quad \displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{(x-3)}{x(x-2)} \bigg]$

$\\\quad\longrightarrow\quad \dfrac{1-3}{1(1-2)}$

$\\\quad\longrightarrow\quad \dfrac{-2}{-1}$

$\\\quad\longrightarrow\quad 2$

$\\\quad\therefore\quad \boxed{\displaystyle \lim_{x \to 1} \rm \bigg[\dfrac{(x^{2} -4x+3}{x(x -1)(x-2)} \bigg] = 2 }$

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