How do you find #\lim _ { x \rightarrow \infty } \sqrt { x ^ { 2} + 1x - 4} - x#?

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Answer 1 · 2017-03-19T17:24:50

#1/2.#

The Reqd. Limit =#lim_(xtooo)sqrt((x^2+x-4))-x.#

#=lim_(xtooo)(sqrt((x^2+x-4))-x))(sqrt((x^2+x-4))+x)/(sqrt(x^2+x-4)+x)#

#=lim_(xtooo){(x^2+x-4)-x^2}/(sqrt(x^2+x-4)+x)#

#=lim_(xtooo)(x-4)/(sqrt(x^2(1+1/x-4/x^2))+x)#

#=lim_(xtooo)(x-4)/(xsqrt(1+1/x-4/x^2)+x)#

#=lim_(xtooo)(x(1-4/x))/{x(sqrt(1+1/x-4/x^2)+1)#

#=lim_(xtooo) (1-4/x)/(sqrt(1+1/x-4/x^2)+1)#

Here, # xtooorArr1/xto0, and, 1/x^2to0.#

#:." The Reqd. Limit="(1-0)/{sqrt(1+0-0)+1}=1/(1+1)=1/2.#

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