Question

lim [x-2/x^2-x - 1/x^3-3x^2+2x]

x=1 ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

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

If we substitute directly x = 1, we get

$\rm \: = \: \bigg[\dfrac{1 - 2}{ {1}^{2} - 1} - \dfrac{1}{ {1}^{3} - {3(1)}^{2} + 2} \bigg]$

$\rm \: = \: - \infty - \infty$

which is indeterminant form.

So,

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

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

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

$\rm \: = \: \displaystyle\lim_{x \to 1}\rm \bigg[\dfrac{x - 2}{x(x - 1)} - \frac{1}{x[x(x - 1) - 2(x - 1)]} \bigg]$

$\rm \: = \: \displaystyle\lim_{x \to 1}\rm \bigg[\dfrac{x - 2}{x(x - 1)} - \frac{1}{x(x - 1)(x - 2)} \bigg]$

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

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

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

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

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

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

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

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

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

$\rm \: = \: 2$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 1}\rm \bigg[\dfrac{x - 2}{ {x}^{2} - x} - \dfrac{1}{ {x}^{3} - {3x}^{2} + 2x} \bigg] = 2}}$

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More to Know

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1}}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}$

Answer 2

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