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?!
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?!
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.
$\displaystyle \displaystyle f(x) = \begin{cases}2(x + 1)\textrm{ if }x < 3 \\ 4 \textrm{ if }x = 3 \\ x^2 - 1 \textrm{ if }x > 3\end{cases}$
note that $\displaystyle 2 < 3$ ...
$\displaystyle \displaystyle \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 6$
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,
$\displaystyle \lim_{x\to 2^-} f(x)= \lim_{x\to 2^-} 2(x+1)= 2(3)= 6$
$\displaystyle \lim_{x\to 2^+} f(x)= \lim_{x\to 2^+} 2(x+1)= 2(3)= 6$