લક્ષ શોધો: mathop {lim }limits_{x to 1} left[ | vedclass

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

Question

(A) આપેલ લક્ષ $\mathop {\lim }\limits_{x 	o 1} \left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}ight]$ છે.પ્રથમ,છેદના અવયવો પાડીએ:$x^2 - x = x(

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) આપેલ લક્ષ $\mathop {\lim }\limits_{x 	o 1} \left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}ight]$ છે.પ્રથમ,છેદના અવયવો પાડીએ:$x^2 - x = x(x-1)$$x^3 - 3x^2 + 2x = x(x-1)(x-2)$હવે,પદાવલિને ફરીથી લખીએ:$\frac{x-2}{x(x-1)} - \frac{1}{x(x-1)(x-2)} = \frac{(x-2)^2 - 1}{x(x-1)(x-2)}$$= \frac{x^2 - 4x + 4 - 1}{x(x-1)(x-2)} = \frac{x^2 - 4x + 3}{x(x-1)(x-2)}$$= \frac{(x-3)(x-1)}{x(x-1)(x-2)}$$x 
eq 1$ માટે સામાન્ય અવયવ $(x-1)$ ને દૂર કરતા:$= \frac{x-3}{x(x-2)}$હવે,$x 	o 1$ માટે લક્ષની કિંમત મૂકતા:$\mathop {\lim }\limits_{x 	o 1} \frac{x-3}{x(x-2)} = \frac{1-3}{1(1-2)} = \frac{-2}{-1} = 2$.

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

Similar Questions

$\mathop {\lim }\limits_{x \to 1} \frac{{x + {x^2} + ...... + {x^n} - n}}{{x - 1}}$ ની કિંમત શોધો.

$\lim_{x \rightarrow \frac{\pi}{2}} (\tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2}))$ ની કિંમત શોધો.

$\lim _{x \rightarrow 0} \frac{63^x-9^x-7^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=\ldots$.

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

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{x^3}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module