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Im having trouble with proving this.
I know that $$\lim_{x\to-1} |x+2| + |x| = 2$$
but Im having trouble with proving it (using the definition of the limit (with $\epsilon$ and $\delta$ )

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    Something that can make your life muuuuuch easier in this case: since $\;x\to-1\;$ , observe that we can then take, for example, $\;-1.1\le x\le-0.9\;$ , but then $$\;x+2>0\;,\;\;x<0\implies |x+2|+|x|=x+2-x=2\; ...$$ – 2017-01-03

3 Answers 3

0

For $x\in[-2,0]$, i.e. $|x+1|\le1$, the function is identically $2$, so that $|f(x)-2|=0$.

Then for any $\epsilon$, we can choose $\delta=1$.

2

For $x<0,|x|=-x$ and for $x>0, |x|=x$, so let's assume $-2

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    The OP is asking for an "$\epsilon-\delta$ proof". – 2017-01-03
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    @MoebiusCorzer: an $\epsilon-\delta$ proof for the limit of a constant is trivial. – 2017-01-03
0

Fix $\epsilon>0$. We want to find $\delta>0$ such that:

$$\forall x\in\Bbb R: \vert x- (-1)\vert<\delta \implies \Big\vert \vert x+2\vert + \vert x\vert -2\Big\vert <\epsilon$$

It is not restrictive to consider that $\delta<1/2$ because if there exists a $\delta$ satisfying the previous assertion, then any smaller $\delta$ will work, as well.

If we consider $\delta<1/2$, we know that $\vert x-(-1)\vert< \delta$ is equivalent to $-\delta - 1 < x < \delta -1 $, hence $$-2<-\tfrac{3}{2}

Hence, for such $x$:

$$ \Big\vert \vert x+2\vert + \vert x\vert -2\Big\vert = \vert x+ 2 + (-x) - 2\vert = 0$$

Hence, we see that for all $\epsilon>0$, taking any $0<\delta <1/2$ works.