What is the limit of #(sqrt(9x^6 - 6)) / (x^3 + 1)# as x goes to negative infinity?

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Answer 1 · 2015-10-09T14:11:13

Here is another way to write up the solution.

#sqrt(9x^6-6) = sqrt (x^6(9-6/x^6))# for #x != 0#

# = sqrt(x^6)sqrt(9-6/x^6)# for #x != 0#

# = abs(x^3) sqrt(9-6/x^6)# for #x != 0#

And #x^3+1 = # for #x != 0#

For #x < 0#, we have #abs(x^3) = -x^3#

As #xrarr-oo#, we are concerned with negative values far from 0.

Putting all this together we write:

#lim_(xrarr-oo) sqrt(9x^6-6)/(x^3+1) = lim_(xrarr-oo) (abs(x^3) sqrt(9-6/x^6))/(x^3(1+1/x^3))#

# = lim_(xrarr-oo) (-x^3 sqrt(9-6/x^6))/(x^3(1+1/x^3))#

# = lim_(xrarr-oo) (-sqrt(9-6/x^6))/(1+1/x^3)#

# = (-sqrt(9-0))/(1+0)#

# = -3#