lim _{y rightarrow 0} frac{sqrt{1+sqrt{1+y^4} | vedclass

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

Question

(C) લિમિટની ગણતરી કરવા માટે,આપણે અંશનું સંમેયીકરણ કરીએ છીએ: $\lim _{y ightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} 	imes \frac{\sqrt{1+\

Vedclass · Accepted Answer

$\frac{1}{4 \sqrt{2}}$

Vedclass · Answer

(C) લિમિટની ગણતરી કરવા માટે,આપણે અંશનું સંમેયીકરણ કરીએ છીએ: $\lim _{y ightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} 	imes \frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$ $= \lim _{y ightarrow 0} \frac{1+\sqrt{1+y^4}-2}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})} = \lim _{y ightarrow 0} \frac{\sqrt{1+y^4}-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})}$ હવે,બાકી રહેલા વર્ગમૂળ પદનું સંમેયીકરણ કરીએ: $= \lim _{y ightarrow 0} \frac{(\sqrt{1+y^4}-1)(\sqrt{1+y^4}+1)}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$ $= \lim _{y ightarrow 0} \frac{1+y^4-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$ $= \lim _{y ightarrow 0} \frac{y^4}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$ $= \frac{1}{(\sqrt{1+1}+\sqrt{2})(1+1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$

$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} = $

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \left[ \frac{\sqrt{a + x} - \sqrt{a - x}}{x} \right]$ ની કિંમત શોધો.

લક્ષ શોધો: $\mathop {\lim }\limits_{x \to -1} (1+x+x^{2}+ \dots +x^{10})$

$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} \right)^{n(n - 1)}} = $

$\lim \limits_{x}$ ${\rightarrow \frac{1}{\sqrt{2}}} \frac{\sin \left(\cos ^{-1} x\right)-x}{1-\tan \left(\cos ^{-1} x\right)}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module