lim _{y rightarrow 0} frac{sqrt{3+y^3}-sqrt{3 | vedclass

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

Question

(A) ધારો કે $L = \lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}$ અંશનું સંમેયીકરણ કરતા: $L = \lim _{y}$ ${\rightarrow 0} \frac{(\sqrt{3+y^3

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{3}}$

Vedclass · Answer

(A) ધારો કે $L = \lim _{y ightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}$ અંશનું સંમેયીકરણ કરતા: $L = \lim _{y}$ ${ightarrow 0} \frac{(\sqrt{3+y^3}-\sqrt{3})}{y^3} 	imes \frac{(\sqrt{3+y^3}+\sqrt{3})}{(\sqrt{3+y^3}+\sqrt{3})}$ $L = \lim _{y ightarrow 0} \frac{3+y^3-3}{y^3(\sqrt{3+y^3}+\sqrt{3})}$ $L = \lim _{y ightarrow 0} \frac{y^3}{y^3(\sqrt{3+y^3}+\sqrt{3})}$ $L = \lim _{y ightarrow 0} \frac{1}{\sqrt{3+y^3}+\sqrt{3}}$ $y = 0$ મૂકતા: $L = \frac{1}{\sqrt{3+0}+\sqrt{3}} = \frac{1}{\sqrt{3}+\sqrt{3}} = \frac{1}{2 \sqrt{3}}$

$\lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3} = $

Similar Questions

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} - \sqrt {{x^2} - \sqrt {{x^2} - \dots} } } }}{x}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }} = $

$\lim _{x \rightarrow \infty}\left(\frac{3 x^2-2 x+3}{3 x^2+x-2}\right)^{3 x-2} = $

$\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = $

$\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos 3 x}{\sin x \log (1+2 x)}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module