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

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well... find the one-sided limits for the piece-wise function if the \(\bf \lim\limits_{x\to 0^-}\textit{ is equal to }\lim\limits_{x\to 0^+} \\ \quad \\ then\qquad \lim\limits_{x\to 0}\ exist\)
\(\large \lim\limits_{x\to 0}\qquad \begin{cases} 7-x^2&x<0\quad \textit{left side, thus } \lim\limits_{x\to 0^-}\\ 7&x=0\\ 10x+7&x>0\quad \textit{right side, thus } \lim\limits_{x\to 0^+} \end{cases}\) 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

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very simple if you simply set x = 0 for those equations btw
i dont get it @jdoe0001 can u help me step by step?
so would the answer be 7? @jdoe0001
.. how did you get 7 though?
i set replaced x with 0 @jdoe0001
am i right or wrong? @jdoe0001
well... is a piece-wise, so there are 3 equations there.. so... which one did you set to 0?
all of them @jdoe0001
am i wrong is the answer not 7? @jdoe0001
\(\bf \lim\limits_{x\to 0^-}\quad 7-x^2=?\impliedby \textit{left side limit} \\ \quad \\ \lim\limits_{x\to 0^+}\quad 10x+7=?\impliedby \textit{right side limit} \) say... what would you get for those two?
i would get 7 @jdoe0001
ok... so, the left-sided limit gives 7 the right-sided limit gives 7 then \(\bf \lim\limits_{x\to 0} \implies 7\) :)
can u help me with another? @jdoe0001
sure, post anew, more eyes :)
thus if I dunno, someone else may help

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