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- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

ok I chose

*being false, since a limit does not exist if f(x) is different coming from $\pm$***c**
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- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 2,678

ok I chose **c** being false, since a limit does not exist if f(x) is different coming from $\pm$

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- #2

- Feb 5, 2013

- 1,382

You need to look at left-hand and right-hand limits. I don't see where "$\pm$" comes into it.

- Jan 29, 2012

- 1,151

Yes, Karush, "c" is the only one that is false.

a) [tex]\lim_{x\to 2} f(x)[/tex] exists.

True. The limit is 2. (f(2)= 1 so f is NOT continuous there.)

b) [tex]\lim_{x\to 3} f(x)[/tex] exists.

True. The limit is 5. (Further f(3)= 5 so f is continuous there.)

c) [tex]\lim_{x\to 4} f(x)[/tex] exists.

False. The "limit from the left", [tex]\lim_{x\to 4^-} f(x)[/tex], is 2 while the "limit from the right", [tex]\lim_{x\to 4^+} f(x)[/tex], is 4. Since the two one-sided limits are not the same the limit itself does not exist.

d) [tex]\lim_{x\to 5} f(x)[/tex] exists.

True. The limit is 6. (Further f(5)= 6 so f is continuous there.)

e) f is continuous at x= 3.

True. As I said in (b), [tex]\lim_{x\to 3} f(x)[/tex] and f(3) both exist and are equal.

- Thread starter
- #4

- Jan 31, 2012

- 2,678

mahalo everyone the comments really increase the insight on these.