mathop {lim}limits_{x to 1} left[ {left[ {fra | vedclass

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

Question

(B) Let $f(x) = \frac{4}{{{x^2} - \frac{1}{x}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x}}{{{x^3} - 1}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $f(x) = \frac{4}{{{x^2} - \frac{1}{x}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x}}{{{x^3} - 1}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x + 1 - 3x + {x^2}}}{{{x^3} - 1}} = \frac{{{x^2} + x + 1}}{{(x - 1)({x^2} + x + 1)}} = \frac{1}{{x - 1}}$.Then,the first term is $[f(x)]^{-1} = x - 1$.The second term is $\frac{{3({x^4} - 1)}}{{{x^3} - \frac{1}{x}}} = \frac{{3x({x^4} - 1)}}{{{x^4} - 1}} = 3x$.Thus,the expression becomes $\mathop {\lim}\limits_{x 	o 1} [(x - 1) + 3x] = \mathop {\lim}\limits_{x 	o 1} (4x - 1) = 4(1) - 1 = 3$.

$\mathop {\lim}\limits_{x \to 1} \left[ {\left[ {\frac{4}{{{x^2} - {x^{ - 1}}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}}} \right]^{ - 1} + \frac{{3 \cdot ({x^4} - 1)}}{{{x^3} - {x^{ - 1}}}}} \right] = $

Similar Questions

If $\mathop {\lim }\limits_{x \to 2} \frac{{{x^n} - {2^n}}}{{x - 2}} = 80$,where $n$ is a positive integer,then $n = $

If $[x]$ is the greatest integer function,then $\lim _{x \rightarrow 3^{-}} \frac{(3-|x|+\sin |3-x|) \cos [9-3 x]}{|3-x|[3 x-9]} = $

If $f: R \rightarrow R$ is defined by $f(x) = [x-3] + |x-4|$ for $x \in R$,then $\lim_{x \rightarrow 3^{-}} f(x)$ is equal to

$\lim _{x \rightarrow 0} \frac{\sin x \sin ^{-1} x}{x^2}$ is equal to

Find the limit: $\mathop {\lim }\limits_{x \to 3} [x(x+1)]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module