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How do you find the limit #(x^3-6x-2)/(x^3-4)# as #x->2#?

Answer 1

#-3/2#

The first thing you have to do is to #color (red) (plug)# in the value of #x# to test if it doesn't give you an indeterminate form. #0/0# and #oo/oo# are some form of indetermination. I think they are the most popular but we have others as well. If you find some indeterminate form, you will have to use some formulas or tricks but if not, then you are lucky and your answer will be the value you just found.

For this one I can already see that you won't get an indeterminate form because I mentally plugged the value of #x# in the denominator, so I'll spare you some time and not talk about the indeterminate form. You can still ask about it if you want to :)

#lim_(x->2)(x^3-6x-2)/(x^3-4)#

So, plugging #2# in #(x^3-6x-2)/(x^3-4)=((2)^3-6(2)-2)/((2)^3-4)=-3/2#

Hope this helps :)

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Answer 2

Aurora Alvarado

To find the limit of (x^3-6x-2)/(x^3-4) as x approaches 2, we substitute 2 into the expression:

(2^3-6(2)-2)/(2^3-4)

Simplifying this expression gives us:

(-2)/(8-4)

Which further simplifies to:

-2/4

And finally, the limit is:

-1/2

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit #(x^3-6x-2)/(x^3-4)# as #x-&gt;2#?

How do you find the limit #(x^3-6x-2)/(x^3-4)# as #x->2#?