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

Answer 1

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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Answer 2

To find ( \lim_{x \to 2} \frac{x^3 - 6x - 2}{x^3 - 4x} ), we can apply L'Hôpital's Rule.

  1. Evaluate the limit directly: Substituting ( x = 2 ) into the expression, we get: [ \frac{2^3 - 6(2) - 2}{2^3 - 4(2)} = \frac{8 - 12 - 2}{8 - 8} = \frac{-6}{0} ]

  2. As the denominator approaches zero, we can use L'Hôpital's Rule: Differentiate the numerator and denominator separately: [ \lim_{x \to 2} \frac{d}{dx}(x^3 - 6x - 2) \Big/ \frac{d}{dx}(x^3 - 4x) ] [ = \lim_{x \to 2} \frac{3x^2 - 6}{3x^2 - 4} ]

  3. Now, evaluate the limit again by substituting ( x = 2 ): [ \lim_{x \to 2} \frac{3(2)^2 - 6}{3(2)^2 - 4} = \frac{3(4) - 6}{3(4) - 4} = \frac{6}{8} = \frac{3}{4} ]

Therefore, ( \lim_{x \to 2} \frac{x^3 - 6x - 2}{x^3 - 4x} = \frac{3}{4} ).

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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.

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.

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.

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.

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