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) माना $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) माना $f(x) = \frac{4}{{{x^2} - \frac{1}{x}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x}}{{{x^3} - 1}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x + 1 - 3x + {x^2}}}{{{x^3} - 1}} = \frac{{{x^2} + x + 1}}{{(x - 1)({x^2} + x + 1)}} = \frac{1}{{x - 1}}$.अतः,पहला पद $[f(x)]^{-1} = x - 1$ है।दूसरा पद $\frac{{3({x^4} - 1)}}{{{x^3} - \frac{1}{x}}} = \frac{{3x({x^4} - 1)}}{{{x^4} - 1}} = 3x$ है।इस प्रकार,व्यंजक $\mathop {\lim}\limits_{x 	o 1} [(x - 1) + 3x] = \mathop {\lim}\limits_{x 	o 1} (4x - 1) = 4(1) - 1 = 3$ हो जाता है।

$\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

जब $x \rightarrow \infty$ हो, तो $\sum_{n=1}^{1000} (-1)^{n} x^{n}$ की सीमा क्या होगी?

$\lim _{x \rightarrow 0} \frac{2x}{|x|+x^2} = $

यदि $\mathop {\lim }\limits_{x \to a} \frac{{{x^9} + {a^9}}}{{x + a}} = 9$ है,तो $a = $

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+a x+b}-x\right]$ जहाँ $a < 0 < b$.

दिए गए सीमा (limit) का मूल्यांकन करें: $\mathop {\lim }\limits_{x \to 0} \frac{ax + x \cos x}{b \sin x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module