HOW TO SOLVE THIS LIMIT

[Guest]

The limit is on the image

[Guest]

Well, the first thing I would try is to set x= 2, y= 1. Oh, no! Both numerator and denominator are 0!
(If neither were 0, the value of the fraction would be the limit. If the numerator were 0 and the denominator not, the limit would be 0. If the denominator were 0 and the numerator not, the limit would not exist.)

But the fact that the numerator and denominator are 0 at x= 2, y= 1 means that they can be factored!
The denominator is easy: [tex]x^2y^2- 4= (xy)^2- 4= (xy- 2)(xy+ 2)[/tex]. For the numerator, I would use the fact that [tex](a- b)(a^2+ ab+ b^2)= a^3- b^3[/tex].

Here, [tex]a= \sqrt[3]{xy+ 6}[/tex] and [tex]b= 2[/tex] so [tex](\sqrt[3]{xy+ 6}+ 2)((\sqrt[3]{xy+ 6}^2+ 2\sqrt[3]{xy+ 6}+ 4)[/tex] so multiplying both numerator and denominator by [tex](\sqrt[3]{xy+ 6})^2+ 2\sqrt[3]{xy+ 6}+ 4[/tex] gives [tex]\frac{(xy- 2)}{(xy-2)(xy+ 2)(\sqrt[3]{xy+ 6}+ 2\sqrt[3]{xy+ 6}+ 4)}[/tex].

As long as xy is not 2, the "xy- 2" terms cancel, leaving [tex]\frac{1}{(xy+ 2)(\sqrt[3]{xy+ 6})^2+ 2\sqrt[3]{xy+ 6}+ 4}[/tex]. Now, taking the limit as x goes to 2 and y goes to 1 is just setting x= 2, y= 1 in that.