Find the limit: mathop {lim }limits_{x to 1} | vedclass

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

Question

(A) We are given the limit $\mathop {\lim }\limits_{x 	o 1} \left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}ight]$.First,we factorize the deno

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) We are given the limit $\mathop {\lim }\limits_{x 	o 1} \left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}ight]$.First,we factorize the denominators:$x^2 - x = x(x-1)$$x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x-1)(x-2)$Now,we rewrite the expression:$\frac{x-2}{x(x-1)} - \frac{1}{x(x-1)(x-2)} = \frac{(x-2)^2 - 1}{x(x-1)(x-2)}$$= \frac{x^2 - 4x + 4 - 1}{x(x-1)(x-2)} = \frac{x^2 - 4x + 3}{x(x-1)(x-2)}$$= \frac{(x-3)(x-1)}{x(x-1)(x-2)}$Canceling the common factor $(x-1)$ for $x 
eq 1$:$= \frac{x-3}{x(x-2)}$Now,evaluating the limit as $x 	o 1$:$\mathop {\lim }\limits_{x 	o 1} \frac{x-3}{x(x-2)} = \frac{1-3}{1(1-2)} = \frac{-2}{-1} = 2$.

Find the limit: $\mathop {\lim }\limits_{x \to 1} \left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}\right]$.

Similar Questions

Evaluate the limit: $\lim _{n \rightarrow \infty} \frac{A+e^{n x}}{x+A e^{n x}}$

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}$ is

$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = $

$\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

$\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)}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module