Lim where f (x) = 2 (x+1) if x <3

x->2 4 if x=3

x^2 - 1 if x> 3

please tell me you have to write out what the left and right limits are to find out if they are equal?

:(

left limit is 6 and right is 3 so the answer is DNE right?!

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- Sep 16th 2011, 12:55 PMhabibixoxLimits of piecewise function
Lim where f (x) = 2 (x+1) if x <3

x->2 4 if x=3

x^2 - 1 if x> 3

please tell me you have to write out what the left and right limits are to find out if they are equal?

:(

left limit is 6 and right is 3 so the answer is DNE right?! - Sep 16th 2011, 12:59 PMskeeterRe: Limits of piecewise function
If $\displaystyle \lim_{x \to c^-} f(x) \ne \lim_{x \to c^+} f(x)$ , then $\displaystyle \lim_{x \to c} f(x)$ does not exist.

- Sep 16th 2011, 09:02 PMProve ItRe: Limits of piecewise function
- Sep 17th 2011, 04:26 AMHallsofIvyRe: Limits of piecewise function
If, in fact, you are trying to find the limit at 2 then you can ignore everything that is not close to 2. In particular, f(x)= 2(x+ 1) for x "anywhere near 2" (close to 2 is certainly less than 3). You do NOT use "$\displaystyle x^2- 1$" to evaluate since that is true only for x> 3.

- Sep 17th 2011, 08:21 AMhabibixoxRe: Limits of piecewise function
yes

- Sep 17th 2011, 08:22 AMhabibixoxRe: Limits of piecewise function
why wouldnt you test out the limits?

- Sep 17th 2011, 08:45 AMhabibixoxRe: Limits of piecewise function
would i be completely wrong if i showed what i did at first on a quiz?

i showed how the right limit doesn't = the left limit so i wrote in conclusion the limit doesn't exist because 3 is not equal to 6 - Sep 17th 2011, 11:05 AMskeeterRe: Limits of piecewise function
$\displaystyle \displaystyle f(x) = \begin{cases}2(x + 1)\textrm{ if }x < 3 \\ 4 \textrm{ if }x = 3 \\ x^2 - 1 \textrm{ if }x > 3\end{cases}$

note that $\displaystyle 2 < 3$ ...

$\displaystyle \displaystyle \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 6$ - Sep 17th 2011, 02:40 PMHallsofIvyRe: Limits of piecewise function
Well, you are not completely wrong to show it now, but you certainly were wrong to do that on the test.

Since the limit is at x= 2, not any where near 3,

$\displaystyle \lim_{x\to 2^-} f(x)= \lim_{x\to 2^-} 2(x+1)= 2(3)= 6$

$\displaystyle \lim_{x\to 2^+} f(x)= \lim_{x\to 2^+} 2(x+1)= 2(3)= 6$