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

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Question

(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,we multiply by $\frac{\sqrt

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,we multiply by $\frac{\sqrt{1+x^2}+\sqrt{1-x+x^2}}{\sqrt{1+x^2}+\sqrt{1-x+x^2}}$:$= \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 $\lim _{x ightarrow 0} \frac{a^x-1}{x} = \log _e a$,we get:$= \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}$.

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

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