How to solve limit of this function?

Question

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

I have tried: 1) substitution of $x$ by $1+\epsilon$(this gets long) 2) factorizing and simplifying

I have to be rigorous and not supposed to use L'Hopital

Answer 1

It's $$\lim_{x\rightarrow1^+}\frac{(x-1)(x-2)}{x^2(x-1)^4}=\lim_{x\rightarrow1^+}\frac{x-2}{x^2(x-1)^3}=-\infty$$ and it's $$\lim_{x\rightarrow1^-}\frac{(x-1)(x-2)}{x^2(x-1)^4}=\lim_{x\rightarrow1^-}\frac{x-2}{x^2(x-1)^3}=+\infty$$

Answer 2

Hint: $$\frac{x^2-3x+2} {(x^3-2x^2+x)^2} = \frac{(x-1)(x-2)}{(x (x-1)^2)^2} = \frac {x-2} {x^4(x-1)}$$

Thus the limit is determined by the behaviour of $\frac 1 {x-1}$ near $1$, which in turn is determined by the behaviour of $\frac 1 x$ near $0$.

Answer 3

In $\lim_{x\to 1} \frac{x^2 -3x+2}{(x^3-2x^2+x)^2} $, let $x = y+1$.

Then

$\begin{array}\\ \frac{x^2 -3x+2}{(x^3-2x^2+x)^2} &=\frac{y(y-1)}{(y+1)^2(x^2-2x+1)^2}\\ &=\frac{y(y-1)}{(y+1)^2((x-1)^2)^2}\\ &=\frac{y(y-1)}{(y+1)^2y^4}\\ &=\frac{y-1}{(y+1)^2y^3}\\ \end{array} $

so the limit is $\pm\infty$ depending how $y \to 0$.