Question

How do you determine the limit of `(x)/sqrt(x^2-x)` as x approaches infinity?

Answer 1

`lim_(xrarroo)x/sqrt(x^2-x) = 1` Bonus: `lim_(xrarr-oo)x/sqrt(x^2-x) = -1`

Explanation:

Note that, for all `x` other than `0` (which we are not interested for limits at infinity), we have

`sqrt(x^2-x) = sqrt(x^2(1-1/x)) = sqrt(x^2)sqrt(1-1/x)`

Note also that `sqrt(x^2) = absx = {(x,"if", x >= 0),(-x,"if",x < 0):}`

So we get

`lim_(xrarroo)x/sqrt(x^2-x) = lim_(xrarroo)x/(sqrt(x^2)sqrt(1-1/x)`

`= lim_(xrarroo)x/(xsqrt(1-1/x)` `" "` (As `xrarroo` we are only interested in positive values of `x`.)

`= lim_(xrarroo)1/(sqrt(1-1/x)) = 1/sqrt(1-0) = 1`

Bonus

`lim_(xrarr-oo)x/sqrt(x^2-x) = lim_(xrarr-oo)x/(sqrt(x^2)sqrt(1-1/x)`

`= lim_(xrarr-oo)x/(-xsqrt(1-1/x)` `" "` (As `xrarr-oo` we are only interested in negative values of `x`.)

`= lim_(xrarroo)1/(-sqrt(1-1/x)) = 1/(-sqrt(1-0)) = -1`