# Thread: Limits of piecewise function

1. ## Limits 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?!

2. ## Re: Limits of piecewise function

If $\lim_{x \to c^-} f(x) \ne \lim_{x \to c^+} f(x)$ , then $\lim_{x \to c} f(x)$ does not exist.

3. ## Re: Limits of piecewise function

Originally Posted by habibixox
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?!
Is this what you are trying to evaluate?

$\displaystyle \lim_{x \to 2} f(x)$ where $\displaystyle f(x) = \begin{cases}2(x + 1)\textrm{ if }x < 3 \\ 4 \textrm{ if }x = 3 \\ x^2 - 1 \textrm{ if }x > 3\end{cases}$

4. ## Re: 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 " $x^2- 1$" to evaluate since that is true only for x> 3.

yes

6. ## Re: Limits of piecewise function

why wouldnt you test out the limits?

7. ## Re: 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

8. ## Re: Limits of piecewise function

$\displaystyle f(x) = \begin{cases}2(x + 1)\textrm{ if }x < 3 \\ 4 \textrm{ if }x = 3 \\ x^2 - 1 \textrm{ if }x > 3\end{cases}$

note that $2 < 3$ ...

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

9. ## Re: Limits of piecewise function

Originally Posted by habibixox
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
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,
$\lim_{x\to 2^-} f(x)= \lim_{x\to 2^-} 2(x+1)= 2(3)= 6$
$\lim_{x\to 2^+} f(x)= \lim_{x\to 2^+} 2(x+1)= 2(3)= 6$