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How do you find #lim (x^3+4x+8)/(2x^3-2)# as #x->1^+# using l'Hospital's Rule or otherwise?

Answer 1

Luke Alvarez

The limit does not exist (it is infinite).

L'Hôpital's rule cannot be used as the limit is not of an indeterminate form #0/0# or #oo/oo#

The limit does not exist (it is infinite).

This is the graph of the function #(x^3+4x+4)/(2x^3-2)#

graph{(x^3+4x+4)/(2x^3-2) [-10, 10, -25, 25]}

As #x rarr 1^+ => (x^3+4x+4)/(2x^3-2) rarr oo #

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Answer 2

Charles Arocha

To find lim (x^3+4x+8)/(2x^3-2) as x approaches 1 from the right using l'Hôpital's Rule, differentiate the numerator and the denominator separately with respect to x, then evaluate the limit of the resulting expressions as x approaches 1 from the right.

Differentiate the numerator: f'(x) = 3x^2 + 4
Differentiate the denominator: g'(x) = 6x^2
Substitute x = 1 into the derivatives: f'(1) = 3(1)^2 + 4 = 7 g'(1) = 6(1)^2 = 6
Evaluate the limit of the ratio of the derivatives: lim (x->1^+) f'(x)/g'(x) = lim (x->1^+) (3x^2 + 4)/(6x^2) = 7/6

So, lim (x^3+4x+8)/(2x^3-2) as x approaches 1 from the right is 7/6.

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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 #lim (x^3+4x+8)/(2x^3-2)# as #x-&gt;1^+# using l'Hospital's Rule or otherwise?

How do you find #lim (x^3+4x+8)/(2x^3-2)# as #x->1^+# using l'Hospital's Rule or otherwise?