Question

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

Explanation:

Step1: Factorize numerator and denominator

The numerator $x^{2}-3x + 2=(x - 1)(x - 2)$ and the denominator $x^{2}-2x + 1=(x - 1)^{2}$. So the limit becomes $\lim_{x
ightarrow1^{+}}\frac{(x - 1)(x - 2)}{(x - 1)^{2}}$.

Step2: Simplify the function

Cancel out the common factor $(x - 1)$ (since $x
eq1$ when taking the limit), we get $\lim_{x
ightarrow1^{+}}\frac{x - 2}{x - 1}$.

Step3: Analyze the one - sided limit

As $x
ightarrow1^{+}$, the numerator approaches $1-2=-1$ and the denominator approaches $0^{+}$. So $\lim_{x
ightarrow1^{+}}\frac{x - 2}{x - 1}=-\infty$.

Answer:

$-\infty$