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lim x (approach to 4) (3x+square root x-4).

If I plug number for x=4, I get 12. But I am concern about square root. Do I need find one-sided limit from x approach to 4+ and x approach to 4-. Or 12 will be the correct answer?

Thanks, everyone

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- Thread starter Strelka
- Start date

- #1

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- 0

lim x (approach to 4) (3x+square root x-4).

If I plug number for x=4, I get 12. But I am concern about square root. Do I need find one-sided limit from x approach to 4+ and x approach to 4-. Or 12 will be the correct answer?

Thanks, everyone

- #2

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If it's not specified,then the superior limit is understood.

Daniel.

Daniel.

- #3

mathman

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- #4

HallsofIvy

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Homework Helper

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- #5

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We both assumed the limit & the function were defined on the reals (real intervals).

Daniel.

Daniel.

- #6

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[tex]\boxed{\lim_{x \rightarrow 4} 3x + \sqrt{x - 4} = 12}[/tex]

- #7

Hurkyl

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When writing your answer, you can always say the limit doesn't exist, and then write down what the one-sided limit is.

P.S. dex: I know everyone knows what you meant, but "limit superior" (written "lim sup") means something else.

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- #9

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Thanks guys a lot

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