Find the limit: mathop {lim }limits_{x to 2} | vedclass

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

Question

(D) We are given the limit $\mathop {\lim }\limits_{x 	o 2} \frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}$.First,we evaluate the function at $x = 2$:$\frac{2^{2}

Vedclass · Accepted Answer

Does not exist

Vedclass · Answer

(D) We are given the limit $\mathop {\lim }\limits_{x 	o 2} \frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}$.First,we evaluate the function at $x = 2$:$\frac{2^{2}-4}{2^{3}-4(2)^{2}+4(2)} = \frac{4-4}{8-16+8} = \frac{0}{0}$.Since it is an indeterminate form,we simplify the expression by factoring:$\frac{x^{2}-4}{x(x^{2}-4x+4)} = \frac{(x-2)(x+2)}{x(x-2)^{2}}$.Canceling the common factor $(x-2)$,we get:$\mathop {\lim }\limits_{x 	o 2} \frac{x+2}{x(x-2)}$.As $x 	o 2$,the numerator approaches $4$ and the denominator approaches $2(0) = 0$.Since the limit results in a non-zero constant divided by zero,the limit does not exist (it approaches infinity).

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}\right]$

Similar Questions

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

$\lim _{x \rightarrow 0} \frac{2 \tan x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 \tan x+1}-\sqrt{3 \tan ^2 x+\sin x+1}} = $

Evaluate the given limit: $\mathop {\lim }\limits_{z \to 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

$\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3} = $

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} (\text{cosec}\,x - \cot x)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module