Question

What is the limit of `(-2x+1)/sqrt(x^2 +x)` as x goes to infinity?

Answer 1

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

Explanation:

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

`= (x(-2+1/x))/(absxsqrt(1+1/x))`

When we find the limit as `xrarroo`, we are concerned with positive values of `x`, so we have:

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

`= lim_(xrarroo) (x(-2+1/x))/(xsqrt(1+1/x))`

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

For limit as `xrarr-oo`

we use the fact the for negative values of `x`

`sqrt(x^2) = absx = -x`, to get

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

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

Here is the graph, so we can see the two horizontal asymptotes.

You can zoom in and out and drag the graph using a mouse.

graph{(-2x+1)/sqrt(x^2 +x) [-25.9, 31.81, -14.4, 14.46]}