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, $xtooorArr1/xto0, and, 1/x^2to0.$

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

Enjoy Maths.!

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