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

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Answer 1 · 2016-05-05T18:54:27

The limit is #1/2#. For "how", please see the explanation section below.

In investigating a limit as #x# increases without bounds, we are not concerned about what happens at #x=0#.

We note that for all #x != 0#,

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

# = (sqrt(x^2)sqrt(1-1/x^2))/(sqrt(x^2)sqrt(4-1/x))#

Note also that #sqrt(x^2) = absx#, so for positive #x#, we have #sqrt(x^2) = x#.

(For negative #x#, we get #sqrt(x^2) = -x#)

We continue:

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

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

Evaluating the limit as #xrarroo#, we get

#lim_(xrarroo)sqrt(x^2-1)/sqrt(4x^2+x) = sqrt(1-0)/sqrt(4+0) = 1/2#