lim _{x rightarrow 2} frac{sqrt[3]{6+x}-sqrt[ | vedclass

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Question

(A) माना $L = \lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2}$ है।सर्वसमिका $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ का उपयोग करते हुए,जहा

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(A) माना $L = \lim _{x ightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2}$ है।सर्वसमिका $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ का उपयोग करते हुए,जहाँ $a = \sqrt[3]{6+x}$ और $b = \sqrt[3]{10-x}$ है:$L = \lim _{x ightarrow 2} \frac{(6+x) - (10-x)}{(x-2)((6+x)^{2/3} + (6+x)^{1/3}(10-x)^{1/3} + (10-x)^{2/3})}$$L = \lim _{x ightarrow 2} \frac{2x - 4}{(x-2)((6+x)^{2/3} + (6+x)^{1/3}(10-x)^{1/3} + (10-x)^{2/3})}$$L = \lim _{x ightarrow 2} \frac{2(x-2)}{(x-2)((6+x)^{2/3} + (6+x)^{1/3}(10-x)^{1/3} + (10-x)^{2/3})}$$L = \frac{2}{8^{2/3} + (8 	imes 8)^{1/3} + 8^{2/3}} = \frac{2}{4 + 4 + 4} = \frac{2}{12} = \frac{1}{6}$.

$\lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2} = $

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सीमा ज्ञात कीजिए: $\mathop {\lim }\limits_{x \to 1} \left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}\right]$.

मान लीजिए $x_{n}=\left(1-\frac{1}{3}\right)^{2}\left(1-\frac{1}{6}\right)^{2}\left(1-\frac{1}{10}\right)^{2} \ldots \left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2}, n \geq 2$ है। तो, $\lim _{n \rightarrow \infty} x_{n}$ का मान ज्ञात कीजिए।

सीमा ज्ञात कीजिए: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{3}-2 x^{2}}{x^{2}-5 x+6}\right]$

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