mathop {lim }limits_{x to 0} frac{{{{(27 + x) | vedclass

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

Question

(C) ધારો કે $L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{(27 + x)}^{\frac{1}{3}}} - 3}}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$$x = 0$ મૂકતા,આપણને $\frac

Vedclass · Accepted Answer

$-\frac{1}{6}$

Vedclass · Answer

(C) ધારો કે $L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{(27 + x)}^{\frac{1}{3}}} - 3}}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$$x = 0$ મૂકતા,આપણને $\frac{0}{0}$ અનિશ્ચિત સ્વરૂપ મળે છે.એલ-હોસ્પિટલના નિયમનો ઉપયોગ કરતા:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}({{(27 + x)}^{\frac{1}{3}}} - 3)}}{{\frac{d}{{dx}}(9 - {{(27 + x)}^{\frac{2}{3}}})}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{3}{{(27 + x)}^{ - \frac{2}{3}}}}}{{ - \frac{2}{3}{{(27 + x)}^{ - \frac{1}{3}}}}}$$L = \mathop {\lim }\limits_{x 	o 0} -\frac{1}{2} {(27 + x)}^{ - \frac{1}{3}}$$L = -\frac{1}{2} 	imes \frac{1}{3} = -\frac{1}{6}$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(27 + x)}^{\frac{1}{3}}}} - 3}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$ ની કિંમત શોધો.

Similar Questions

$\lim _{x \rightarrow 1} \frac{\log x}{1-x} = $

$\mathop {\lim }\limits_{x \to \alpha } \frac{{\sin x - \sin \alpha }}{{x - \alpha }} = $

ધારો કે $f(2) = 4$ અને $f'(2) = 4$,તો $\mathop {\lim }\limits_{x \to 2} \,\frac{{xf(2) - 2f(x)}}{{x - 2}}$ ની કિંમત શોધો.

$\mathop {Limit}\limits_{x \to 0^+} \frac{1}{x\sqrt{x}} \left( a \tan^{-1} \frac{\sqrt{x}}{a} - b \tan^{-1} \frac{\sqrt{x}}{b} \right)$ ની કિંમત કેટલી થાય?

$\mathop {\lim }\limits_{x \to \pi /4} \frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module