mathop {lim }limits_{x to infty } frac{{sqrt | vedclass

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Question

(A) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{\sqrt {{x^2} + {a^2}} - \sqrt {{x^2} + {b^2}} }}{{\sqrt {{x^2} + {c^2}} - \sq

Vedclass · Accepted Answer

$\frac{{{a^2} - {b^2}}}{{{c^2} - {d^2}}}$

Vedclass · Answer

(A) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{\sqrt {{x^2} + {a^2}} - \sqrt {{x^2} + {b^2}} }}{{\sqrt {{x^2} + {c^2}} - \sqrt {{x^2} + {d^2}} }}$,we rationalize the numerator and the denominator: Multiply the numerator and denominator by their respective conjugates: $= \mathop {\lim }\limits_{x 	o \infty } \frac{(\sqrt {x^2 + a^2} - \sqrt {x^2 + b^2})(\sqrt {x^2 + a^2} + \sqrt {x^2 + b^2})}{(\sqrt {x^2 + c^2} - \sqrt {x^2 + d^2})(\sqrt {x^2 + c^2} + \sqrt {x^2 + d^2})} 	imes \frac{(\sqrt {x^2 + c^2} + \sqrt {x^2 + d^2})}{(\sqrt {x^2 + a^2} + \sqrt {x^2 + b^2})} $$= \mathop {\lim }\limits_{x 	o \infty } \frac{(x^2 + a^2) - (x^2 + b^2)}{(x^2 + c^2) - (x^2 + d^2)} 	imes \frac{\sqrt {x^2 + c^2} + \sqrt {x^2 + d^2}}{\sqrt {x^2 + a^2} + \sqrt {x^2 + b^2}} $$= \mathop {\lim }\limits_{x 	o \infty } \frac{a^2 - b^2}{c^2 - d^2} 	imes \frac{x\sqrt {1 + c^2/x^2} + x\sqrt {1 + d^2/x^2}}{x\sqrt {1 + a^2/x^2} + x\sqrt {1 + b^2/x^2}} $As $x 	o \infty$,the terms $a^2/x^2, b^2/x^2, c^2/x^2, d^2/x^2$ approach $0$. $= \frac{a^2 - b^2}{c^2 - d^2} 	imes \frac{1 + 1}{1 + 1} = \frac{a^2 - b^2}{c^2 - d^2}$.

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} + {a^2}} - \sqrt {{x^2} + {b^2}} }}{{\sqrt {{x^2} + {c^2}} - \sqrt {{x^2} + {d^2}} }} = $

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