How do you find the limit of #(x^3 + 1) / (sqrt { 4x^2 + 3} )# as x approaches infinity?

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Answer 1 · 2015-06-20T18:26:51

Use #sqrt(4x^2+3) = sqrt(x^2)sqrt(4+3/x^2)#, for #x != 0# and #=xsqrt(4+3/x^2)# for #x > 0#

This answer assumes that you either do not have l'Hopital's Rule, or you prefer not to use it.

Note: #sqrt(x^2) = abs(x) = { (x, " if " , x >=0),(-x, " if ", x<0) :}#

To find #lim_(xrarroo)#, we are only interested in positive values of #x#, so we can use:

#(x^3 + 1) / (sqrt { 4x^2 + 3} ) = (x^3+1)/sqrt(x^2(4+3/x^2)#

# = (x^3+1)/(sqrt(x^2)sqrt(4+3/x^2))#

# = (x(x^2+1/x))/(xsqrt(4+3/x^2))# (for #x>0#)

# = (x^2+1/x)/sqrt(4+3/x^2)#

Clearly, this last ratio approaches #x^2/2# as #xrarroo#, so the limit is #oo#

#lim_(xrarroo)(x^3 + 1) / (sqrt { 4x^2 + 3} ) = lim_(xrarroo) (x^2+1/x)/sqrt(4+3/x^2) = oo#