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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(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

Similar Questions

$\lim _{x \rightarrow \infty} \frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)\sqrt{(3x+2)^x}}$ is equal to:

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ({x^{1/3}})\ln (1 + 3x)}}{{{{(\tan^{ - 1}\sqrt x )}^2}({e^{5{x^{1/3}}}} - 1)}} = $

$\mathop {\lim }\limits_{x \to 1} [x] = $

$\lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2} = $

$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 - \cos 2(x - 1)} }}{{x - 1}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module