# Thread: [SOLVED] LH and RH limits of a piecewise function

1. ## [SOLVED] LH and RH limits of a piecewise function

When evaluating the continuity of a piecewise function by determining if the LH and RH limits exist and are equal to the function value, how does one go about finding the LH and RH limits of the "pieces"?

eg.

f(apple) =
{ x+2 if x<= 0
{ x-3, if x>0

Should I find the LH and RH limits of x+2 and then find the LH and RH limits of x-3, or should i find the LH lim of x-3 and the RH lim of x+2?

I have a feeling it is the former, but I am unsure.

2. Originally Posted by Noxide
When evaluating the continuity of a piecewise function by determining if the LH and RH limits exist and are equal to the function value, how does one go about finding the LH and RH limits of the "pieces"?

eg.

f(apple) =
{ x+2 if x<= 0
{ x-3, if x>0

Should I find the LH and RH limits of x+2 and then find the LH and RH limits of x-3, or should i find the LH lim of x-3 and the RH lim of x+2?

I have a feeling it is the former, but I am unsure.
... note that (x+2) is f(x) for values of x to the left of 0 and 0.

$\lim_{x \to 0^-} (x+2) = 2$

... note that (x-3) is f(x) for values to the right of 0

$\lim_{x \to 0^+} (x-3) = -3$

left limit $\ne$ right limit

therefore, $\lim_{x \to 0} f(x)$ does not exist.

3. Thank you!

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