How do you find #lim_(xto2) (x^3-6x-2)/(x^3-4x)# using l'Hospital's Rule or otherwise?

1 Answer
Truong-Son N.
Jun 26, 2017

The two-sided limit doesn't exist...

graph{(x^3 - 6x - 2)/(x^3 - 4x) [-4.245, 9.8, -1.71, 5.313]}

Well, you can approach it with some manipulation.

#= lim_(x->2) (x^3 - 4x - 2x - 2)/(x^3 - 4x)#

#= lim_(x->2) cancel((x^3 - 4x)/(x^3 - 4x))^(1) - (2x + 2)/(x^3 - 4x)#

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

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

#= 1 - 2[lim_(x->2) (cancelx)/(cancelx(x^2 - 4)) + 1/(x(x^2 - 4))]#

#= 1 - 2[lim_(x->2) 1/((x+2)(x-2)) + 1/(x(x+2)(x-2))]#

We really can't use L'Hopital's rule here because it's not of the form #0/0# or #oo/oo#, but #1/0#. There's no other obvious manipulation we can do.

So, plugging in #2#, the limit is undefined. From the left, it is #oo#, and from the right, it is #-oo#.

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