mathop {lim }limits_{x to 0} frac{{sqrt {1 - | vedclass

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Question

(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}}$,we rationalize the numerator:$=

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}}$,we rationalize the numerator:$= \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}} 	imes \frac{{\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} }}{{\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} }}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{(1 - {x^2}) - (1 + {x^2})}}{{{x^2}(\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} )}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{ - 2{x^2}}}{{{x^2}(\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} )}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{ - 2}}{{\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} }}$$= \frac{{ - 2}}{{\sqrt {1 - 0} + \sqrt {1 + 0} }} = \frac{{ - 2}}{{1 + 1}} = -1$

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}}$ is equal to

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