How do you find the limit of #x / sqrt (4x^2 + 2x +1) # as x approaches infinity?

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Answer 1 · 2016-11-29T18:26:34

Please see below.

#x / sqrt (4x^2 + 2x +1) = x / (sqrt(x^2)sqrt (4 + 2/x +1/x^2))# #" "# for #x != 0#

#sqrt(x^2) = absx = x# for #x > 0#

#lim_(xrarroo) x/ sqrt (4x^2 + 2x +1) = lim_(xrarroo)x / (xsqrt (4 + 2/x +1/x^2))#

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

# = 1/sqrt(4+0+0) = 1/2#

Bonus material

Since #sqrt(x^2) = -x# for #x < 0#, we have

#lim_(xrarr-oo) x/ sqrt (4x^2 + 2x +1) = lim_(xrarr-oo)x / (-xsqrt (4 + 2/x +1/x^2))#

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

# = -1/sqrt(4+0+0) = - 1/2#

Here is the graph

graph{x / sqrt (4x^2 + 2x +1) [-10, 10, -5, 5]}