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

Answer 1

Because the expression evaluated at #x =2# does not yield an indeterminate form, (e.g. #0/0# or #oo/oo#), one cannot use L'Hospital's rule. The limit is merely the expression evaluated at #x = 2#.

Evaluate the expression at #x = 2#:

#lim_(x to 2) (x^3-6x-2)/(x^3+4) = (2^3-6(2)-2)/(2^3+4)#

The limit is:

#lim_(x to 2) (x^3-6x-2)/(x^3+4) = -1/2#

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Answer 2

Elijah Abram

To find (\lim_{x \to 2} \frac{x^3 - 6x - 2}{x^3 + 4}), we can first try direct substitution. However, substituting (x = 2) gives an indeterminate form of (\frac{0}{12}). To resolve this, we can use L'Hôpital's Rule.

By applying L'Hôpital's Rule, we differentiate the numerator and denominator separately and then evaluate the limit again.

[ \lim_{x \to 2} \frac{x^3 - 6x - 2}{x^3 + 4} = \lim_{x \to 2} \frac{3x^2 - 6}{3x^2} ]

After applying L'Hôpital's Rule once more, we get:

[ = \lim_{x \to 2} \frac{6x}{6x} = 1 ]

So, (\lim_{x \to 2} \frac{x^3 - 6x - 2}{x^3 + 4} = 1).

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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.