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

Question

9889 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-16T14:51:13

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

#(-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]}