Tuesday

July 29, 2014

July 29, 2014

Posted by **Student (UCI)** on Tuesday, February 7, 2012 at 9:24pm.

the the lim x-> 2 f(4x^2-11) = 2

the answer is true, but i do not understand why..because when i plug the numbers in im not getting what it has stated. Please explain thanks:)

- CALC 2A -
**drwls**, Tuesday, February 7, 2012 at 10:07pmWhen x = 2, 4x^2-11 = 5

You have been told that f(x) is continuous at x = 5, and that the value of f(5) = 2.

Therefore the answer is true.

The question is a bit confusing because you have to think of 4x^2-11 as the argument of the f(x) function.

- CALC 2A -
**Student (UCI)**, Wednesday, February 8, 2012 at 11:22amThanks that makes much more sense:)

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