Question

Find the indicated limit, if it exists.
@phi @solomonzelman @jdoe0001 @paki @aaronq @zepdrix @e.mccormick @thomaster

Answer 1

well... find the one-sided limits for the piece-wise function

if the \(\bf \lim\limits_{x\to 0^-}\textit{ is equal to }\lim\limits_{x\to 0^+}
\\ \quad \\
then\qquad \lim\limits_{x\to 0}\ exist\)

Answer 2

\(\large \lim\limits_{x\to 0}\qquad
\begin{cases}
7-x^2&x<0\quad \textit{left side, thus } \lim\limits_{x\to 0^-}\\
7&x=0\\
10x+7&x>0\quad \textit{right side, thus } \lim\limits_{x\to 0^+}
\end{cases}\)

that is, if the left-side of the "limit" matches the right-side of the "limit"
in this case those 2 equations, (notice the limit occurs when "x" approaches 0)
then the double-sided limit exist and is THAT

Answer 3

very simple if you simply set x = 0 for those equations btw

Answer 4

i dont get it @jdoe0001 can u help me step by step?

Answer 5

so would the answer be 7? @jdoe0001

Answer 6

.. how did you get 7 though?

Answer 7

i set replaced x with 0 @jdoe0001

Answer 8

am i right or wrong? @jdoe0001

Answer 9

well... is a piece-wise, so there are 3 equations there.. so... which one did you set to 0?

Answer 10

all of them @jdoe0001

Answer 11

am i wrong is the answer not 7? @jdoe0001

Answer 12

\(\bf \lim\limits_{x\to 0^-}\quad 7-x^2=?\impliedby \textit{left side limit}
\\ \quad \\
\lim\limits_{x\to 0^+}\quad 10x+7=?\impliedby \textit{right side limit}
\)

say... what would you get for those two?

Answer 13

i would get 7 @jdoe0001

Answer 14

ok... so, the left-sided limit gives 7
the right-sided limit gives 7
then \(\bf \lim\limits_{x\to 0} \implies 7\) :)

Answer 15

can u help me with another? @jdoe0001

Answer 16

sure, post anew, more eyes :)

Answer 17

thus if I dunno, someone else may help