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Mth 277
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Question: Find the limit as t -> 2 of ((t^4-2)/(t-2))i + ((t^2-4)/(t^2-2t))j + ((t^2 + 3)e^(t-2))k.
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Your solution:
??????I’m having trouble with this problem. It seems that this vector function is not continuous at the point that the limit is trying to be taken. l’Hospital’s rule can apply to the j component and get a limit of 2 as t 2 and for k we get limit = 7 as t 2. Just not sure what to do for i component???????????????
confidence rating #$&*:
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@& The numerator doesn't approach zero, so l'Hopital's rule doesn't apply. The numerator doesn't approach zero, the denominator does. Whichever side you approach from the numerator and denominator have the same sign, so the limit is +infinity.*@
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& The limit doesn't care a bit what happens at the limiting point.
However if the limit doesn't exist, or approaches infinity, then the function is not continuous at the point.
If the limit does exist, but is not equal to the value of the function at the point, then again the function is not continuous at the point.
In this cae the expression is
(t^4 - 2) / (t - 2).
As t -> 2, the numerator approaches 14, while the denominator approaches zero.
As you approach from the right both numerator and denominator are positive, and as you approach from the left they are both negative. So the limit is +infinity.
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So because the limit of the i component approaches infinity the limit as t -> 2 of ((t^4-2)/(t-2))i + ((t^2-4)/(t^2-2t))j + ((t^2 + 3)e^(t-2))k, is infinity or just the i component??????
@& Very good.
The magnitude of the vector would approach infinity.
The limiting direction would be the `i direction.*@