Rational fractional functions

Compute the limits

Variant 1
\(\displaystyle \lim_{x\to 1}\frac{x^2+2x-3}{x^2-1} \)
Resolution
\(\displaystyle \lim_{x\to 1}\frac{x^2+2x-3}{x^2-1} = \lim_{x\to 1}\frac{(x-1)(x+3)}{(x-1)(x+1)} = \lim_{x\to 1}\frac{x-1}{x-1} \cdot \lim_{x\to 1}\frac{x+3}{x+1} = 1 \cdot \frac42 \)

For completeness, note that in the punctured neighborhood around \(x=1\) the function \(\frac{x-1}{x-1}\) equals 1 everywhere.
Result
The limit equals 2.
Variant 2
\(\displaystyle \lim_{x\to 1}\frac{x^3-3x+2}{x^4-4x+3} \)
Resolution
\(\displaystyle \lim_{x\to 1}\frac{x^3-3x+2}{x^4-4x+3} = \lim_{x\to 1}\frac{(x-1)^2(x+2)}{(x-1)^2(x^2+2x+3)} = \lim_{x\to 1}\frac{x+2}{x^2+2x+3} = \frac36 \)
Result
The limit equals \(\frac12\).
Variant 3
\(\displaystyle \lim_{x\to 2}\frac{(x^2-x-2)^{20}}{(x^3-12x+16)^{10}} \)
Resolution
\(\displaystyle \lim_{x\to 2}\frac{(x^2-x-2)^{20}}{(x^3-12x+16)^{10}}= \lim_{x\to 2}\frac{[(x-2)(x+1)]^{20}}{[(x-2)^2(x+4)]^{10}}= \lim_{x\to 2}\frac{(x+1)^{20}}{(x+4)^{10}}= \frac{3^{20}}{6^{10}} \)
Result
The limit equals \(\left(\frac32\right)^{10}\).
Variant 4
\(\displaystyle \lim_{x\to 1^\pm}\frac{x^2-1}{x^3-x^2-x+1} \)
Resolution
\(\displaystyle \lim_{x\to 1^\pm}\frac{x^2-1}{x^3-x^2-x+1}= \lim_{x\to 1^\pm}\frac{(x-1)(x+1)}{(x-1)^2(x+1)}= \lim_{x\to 1^\pm}\frac1{x-1} \)

\(\displaystyle \lim_{x\to 1^-}\frac1{x-1}=-\infty, \qquad \lim_{x\to 1^+}\frac1{x-1}=\infty \)
Result
The limit from the left equals \(-\infty\); from the right it is \(\infty\).