# CALC 2A

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True or False. If f is continuous at 5 and f(5)=2 and f(4)=3,
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 -

When 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.

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 -

Thanks that makes much more sense:)

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