lim _{x rightarrow 0} frac{1-cos left(x^2+pi( | vedclass

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(C) हमारे पास है,$\lim _{x ightarrow 0} \frac{1-\cos \left(x^2+\pi x+2\piight)}{x^2}$ चूंकि $\cos(2\pi + 	heta) = \cos 	heta$,व्यंजक $\lim _{x \

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$\frac{\pi^2}{2}$

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(C) हमारे पास है,$\lim _{x ightarrow 0} \frac{1-\cos \left(x^2+\pi x+2\piight)}{x^2}$ चूंकि $\cos(2\pi + 	heta) = \cos 	heta$,व्यंजक $\lim _{x ightarrow 0} \frac{1-\cos \left(x^2+\pi xight)}{x^2}$ बन जाता है। सर्वसमिका $1 - \cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$ का उपयोग करने पर: $\lim _{x ightarrow 0} \frac{2 \sin^2 \left(\frac{x^2+\pi x}{2}ight)}{x^2}$ $= 2 \lim _{x}$ ${ightarrow 0} \left[ \frac{\sin \left(\frac{x^2+\pi x}{2}ight)}{\frac{x^2+\pi x}{2}} ight]^2 	imes \left( \frac{x^2+\pi x}{2} ight)^2 	imes \frac{1}{x^2}$ $= 2 	imes 1^2 	imes \lim _{x ightarrow 0} \frac{x^2(x+\pi)^2}{4x^2}$ $= 2 	imes \frac{1}{4} 	imes \pi^2 = \frac{\pi^2}{2}$

$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2} = $

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$\lim _{x \rightarrow 0} x^3 \left\{ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right\} = $

यदि $l = \lim_{x \rightarrow 0} \frac{x}{|x| + x^2}$ है,तो $l$ का मान क्या है?

दिए गए सीमा (limit) का मूल्यांकन करें: $\mathop {\lim }\limits_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$

यदि $n$ एक पूर्णांक है,तो $\mathop {\lim }\limits_{x \to n + 0} (x - [x]) = $

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