lim _{x rightarrow 0} frac{sqrt{1+x^2}-sqrt{1 | vedclass

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(C) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ Rationalizing the numerator: $= \lim _{x}$ ${\rightarro

Vedclass · Accepted Answer

$\frac{1}{\log _e 9}$

Vedclass · Answer

(C) We evaluate the limit: $\lim _{x ightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ Rationalizing the numerator: $= \lim _{x}$ ${ightarrow 0} \frac{(\sqrt{1+x^2}-\sqrt{1-x+x^2})(\sqrt{1+x^2}+\sqrt{1-x+x^2})}{(3^x-1)(\sqrt{1+x^2}+\sqrt{1-x+x^2})}$ $= \lim _{x ightarrow 0} \frac{(1+x^2)-(1-x+x^2)}{(3^x-1)(\sqrt{1+x^2}+\sqrt{1-x+x^2})}$ $= \lim _{x ightarrow 0} \frac{x}{(3^x-1)(\sqrt{1+x^2}+\sqrt{1-x+x^2})}$ $= \lim _{x}$ ${ightarrow 0} \left( \frac{1}{\frac{3^x-1}{x}} ight) 	imes \left( \frac{1}{\sqrt{1+x^2}+\sqrt{1-x+x^2}} ight)$ Using the standard limit $\lim _{x ightarrow 0} \frac{a^x-1}{x} = \log _e a$: $= \frac{1}{\log _e 3} 	imes \frac{1}{\sqrt{1+0}+\sqrt{1-0+0}}$ $= \frac{1}{\log _e 3} 	imes \frac{1}{1+1} = \frac{1}{2 \log _e 3}$ $= \frac{1}{\log _e 3^2} = \frac{1}{\log _e 9}$

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot \mathbf{b} \times \mathbf{c}$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ is equal to

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