mathop {lim }limits_{x to infty } (sqrt {{x^2 | vedclass

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(C) To evaluate the limit,we rationalize the expression: $\mathop {\lim }\limits_{x 	o \infty } (\sqrt {{x^2} + 1} - x) 	imes \frac{{\sqrt {{x^2} +

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(C) To evaluate the limit,we rationalize the expression: $\mathop {\lim }\limits_{x 	o \infty } (\sqrt {{x^2} + 1} - x) 	imes \frac{{\sqrt {{x^2} + 1} + x}}{{\sqrt {{x^2} + 1} + x}}$ $= \mathop {\lim }\limits_{x 	o \infty } \frac{{({x^2} + 1) - {x^2}}}{{\sqrt {{x^2} + 1} + x}}$ $= \mathop {\lim }\limits_{x 	o \infty } \frac{1}{{\sqrt {{x^2} + 1} + x}}$ As $x 	o \infty$,the denominator $\sqrt {{x^2} + 1} + x 	o \infty$. Therefore,$\frac{1}{\infty} = 0$.

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 1} - x)$ is equal to

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