How do you evaluate #[x/(x^2-d^2)^(1/2)] # as x approaches negative infinity?

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Answer 1 · 2016-07-11T14:22:54

-1

#lim_{ x to -oo} [x/(x^2-d^2)^(1/2)]#

#lim_{ x to -oo} sgn(x) [1/(1-(d/x)^2)^(1/2)]# because the denominator is always positive

and
#lim_{ x to -oo} (d/x)^2 = 0#

#implies lim_{ x to -oo} [x/(x^2-d^2)^(1/2)] = -1#

Answer 2 · 2016-07-11T14:26:18

the reqd. limit#=1.#

Let #f(x)=[x/(x^2-d^2)^(1/2)]=[x/{x^2(1-d^2/x^2)}^(1/2)]=[x/(x*{1-(d/x)^2}^(1/2)]]=[cancelx/(cancelx*{1-(d/x)^2}^(1/2)]]=[1/({1-(d/x)^2}^(1/2)]]#

Now, letting, #xrarr -oo#, we have, #d/xrarr0#

Hence, the reqd. limit#=1.#