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Q. $\lim_{x \to {-1}} \frac {x^2+3x+2} {x^2 -4x +3} $

Solution:

$\lim_{x\to-1} \frac{x^{2} +3x +2}{x^{2}+ 4x + 3} = \lim_{x\to-1} \frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+3\right)}$
= $\lim_{x\to-1} \frac{x+2}{x+3} =\frac{-1+2}{-1+3} = \frac{1}{2}$