The value of mathop {lim }limits_{x to - 1} f | vedclass

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(D) To find the limit $\mathop {\lim }\limits_{x 	o - 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 4x + 3}}$,we first factorize the numerator and the denomina

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$1/2$

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(D) To find the limit $\mathop {\lim }\limits_{x 	o - 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 4x + 3}}$,we first factorize the numerator and the denominator.Numerator: $x^2 + 3x + 2 = (x + 1)(x + 2)$Denominator: $x^2 + 4x + 3 = (x + 1)(x + 3)$Now,substitute these into the limit expression:$\mathop {\lim }\limits_{x 	o - 1} \frac{(x + 1)(x + 2)}{(x + 1)(x + 3)}$Since $x 	o - 1$,$x + 1 
eq 0$,we can cancel the common factor $(x + 1)$:$\mathop {\lim }\limits_{x 	o - 1} \frac{x + 2}{x + 3}$Now,evaluate the limit by substituting $x = - 1$:$\frac{- 1 + 2}{- 1 + 3} = \frac{1}{2}$.

The value of $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 4x + 3}}$ is equal to

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