help find indicated limits

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help find indicated limits

Mathematics
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\(\Large\color{slate}{\displaystyle\lim_{x \rightarrow~ -5}f(x)}\) where the function is: \(\LARGE\color{black}{ f(x) = \begin{cases} & x+4,~~~{\large x<-5} \\ & 4-x,~~~{\large x\ge-5} \end{cases} }\)
So, for a two-sided limit \(\Large\color{slate}{\displaystyle\lim_{x \rightarrow~ -5}f(x)}\) to exist, the limit from the right side and the limit from the left side have to be equal to each other. \(\Large\color{slate}{\displaystyle\lim_{x \rightarrow~ -5^+}f(x)=\lim_{x \rightarrow~ -5^-}f(x)}\)
What is our function from the left of -5? (when x is less than -5) it is: x+4 and the function from the right side is 4-x. So this limit: \(\Large\color{slate}{\displaystyle\lim_{x \rightarrow~ -5^+}f(x)=\lim_{x \rightarrow~ -5^-}f(x)}\) would go the following way \(\Large\color{slate}{\displaystyle\lim_{x \rightarrow~ -5^+}(4-x) =\lim_{x \rightarrow~ -5^-}(x+4) }\)
if both sides are not equivalent to each other, then the two sides limit Does Not Exist (DNE). if both sides are equivalent, then this value they are both equal is going to be your answer.
now, do the substitution directly (plug in -5 for x on both sides)
so it does not exist
what about the other question?
Yes, from the right side the limit is 9 and from the left side the limit is -1. Does Not exist is correct
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Do you see the part of this (piece wise) function that goes from the right to the point where x=3?
the function is a broken line at point 3, on x-axis with the lines being with different slope and a point at the top the limit simply doesn't exist.
Em, do you see that segment where the line goes from the right to the point x=3?

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