$$lim_{x \to 1^{+}} \frac{x^{2}-3x + 2}{x^{2}-2x + 1}$$
$$lim_{x \to 1^{+}} \frac{x^{2}-3x + 2}{x^{2}-2x + 1}$$
Answer
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 $\lim_{x\rightarrow1^{+}}\frac{(x - 1)(x - 2)}{(x - 1)^{2}}$.
Step2: Simplify the function
Cancel out the common factor $(x - 1)$ (since $x\neq1$ when taking the limit), we get $\lim_{x\rightarrow1^{+}}\frac{x - 2}{x - 1}$.
Step3: Analyze the one - sided limit
As $x\rightarrow1^{+}$, the numerator approaches $1-2=-1$ and the denominator approaches $0^{+}$. So $\lim_{x\rightarrow1^{+}}\frac{x - 2}{x - 1}=-\infty$.
Answer:
$-\infty$