lim _{x rightarrow 0} frac{sqrt{1+x sin x}-sq | vedclass

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

Question

(D) આપેલ લક્ષ $\lim _{x ightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{	an ^2 2 x}$ છે,જે $\frac{0}{0}$ સ્વરૂપમાં છે. અંશનું સંમેયીકરણ કરતા: $

Vedclass · Accepted Answer

$\frac{3}{16}$

Vedclass · Answer

(D) આપેલ લક્ષ $\lim _{x ightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{	an ^2 2 x}$ છે,જે $\frac{0}{0}$ સ્વરૂપમાં છે. અંશનું સંમેયીકરણ કરતા: $= \lim _{x}$ ${ightarrow 0} \frac{(\sqrt{1+x \sin x}-\sqrt{\cos x})(\sqrt{1+x \sin x}+\sqrt{\cos x})}{	an ^2 2 x (\sqrt{1+x \sin x}+\sqrt{\cos x})}$ $= \lim _{x ightarrow 0} \frac{1+x \sin x-\cos x}{	an ^2 2 x} \cdot \frac{1}{\sqrt{1+x \sin x}+\sqrt{\cos x}} $ $= \lim _{x ightarrow 0} \frac{(1-\cos x)+x \sin x}{	an ^2 2 x} \cdot \frac{1}{2} $ નિત્યસમ $1-\cos x = 2 \sin ^2 \frac{x}{2}$ અને $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ નો ઉપયોગ કરતા: $= \lim _{x}$ ${ightarrow 0} \frac{2 \sin ^2 \frac{x}{2} + 2x \sin \frac{x}{2} \cos \frac{x}{2}}{	an ^2 2 x} \cdot \frac{1}{2}$ $= \lim _{x}$ ${ightarrow 0} \frac{2 \sin ^2 \frac{x}{2} [1 + \frac{x \cos \frac{x}{2}}{\sin \frac{x}{2}}]}{	an ^2 2 x} \cdot \frac{1}{2}$ $= \lim _{x ightarrow 0} \frac{2 \sin ^2 \frac{x}{2} [1 + \frac{x}{	an \frac{x}{2}}]}{	an ^2 2 x} \cdot \frac{1}{2} $ $= \lim _{x ightarrow 0} \frac{2 (\frac{x}{2})^2 [1 + 2 \cdot \frac{x/2}{	an (x/2)}]}{(2x)^2} \cdot \frac{1}{2} $ $= \lim _{x}$ ${ightarrow 0} \frac{2 \cdot \frac{x^2}{4} [1+2]}{4x^2} \cdot \frac{1}{2} = \frac{1/2 \cdot 3}{4} \cdot \frac{1}{2} = \frac{3}{16}$.

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 2 x}=$

Similar Questions

$\lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3} = $

જો $0 < x < y$ હોય,તો $\mathop {\lim }\limits_{n \to \infty } {({y^n} + {x^n})^{1/n}}$ ની કિંમત શું થાય?

$\lim _{x \rightarrow 3 / 2} \frac{\left(4 x^2-6 x\right)\left(4 x^2+6 x+9\right)}{\sqrt[3]{2 x}-\sqrt[3]{3}}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module