To Determine:

Find the limits.

\(\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{x^3-2x^2}\)

Explanation:

To find the limit, we will factorize the numerator and denominator.

\(\lim_{x\rightarrow2^-}\frac{(x-2)(x-1)}{x^2(x-2)}\)

\(\lim_{x\rightarrow2^-}\frac{(x-1)}{x^2}\)

find the limit now

\(\frac{(2-1)}{2^2}=\frac{1}{4}\)

\(=0.25\)

Answer: \(\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{x^3-2x}=0.25\)

Find the limits.

\(\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{x^3-2x^2}\)

Explanation:

To find the limit, we will factorize the numerator and denominator.

\(\lim_{x\rightarrow2^-}\frac{(x-2)(x-1)}{x^2(x-2)}\)

\(\lim_{x\rightarrow2^-}\frac{(x-1)}{x^2}\)

find the limit now

\(\frac{(2-1)}{2^2}=\frac{1}{4}\)

\(=0.25\)

Answer: \(\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{x^3-2x}=0.25\)