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I already have the answer sheet for this problem set, and the proof they give for this problem is somewhat different from mine, so I just wanted someone to check if my proof is valid.

Problem: Let $A(x)=\int_{-2}^{x} f(t)\ \mathrm{d}t$ where $f(t)=-1$ if $t<0$ and $f(t)=1$ if $t \geq 0$. Using $\epsilon$, $\delta$, show that $\lim_{x \rightarrow 0} A(x)$ exists and find its value.

My solution: We wish to prove that there is some $L$ such that, for every $\epsilon>0$ there is a $\delta>0$ such that $\left|A(x)-L\right|<\epsilon$ whenever $0<\vert x\vert<\delta$. Suppose we choose $L=-2$ and $\delta=\epsilon$. Then, for $x=\frac{1}{2}\epsilon<\delta$, we have $$\left|A(x)-L\right|=\left|\int_{-2}^{\frac{1}{2}\epsilon} f(t)\ \mathrm{d}t+2\right|=\left|\int_{-2}^{0} -1\ \mathrm{d}t+\int_{0}^{\frac{1}{2}\epsilon} 1\ \mathrm{d}t+2\right| = \left|\frac{1}{2}\epsilon\right|=\frac{1}{2}\epsilon<\epsilon$$ So the limit exists and we have $\lim_{x \rightarrow 0} A(x)=L=-2$.

Thanks for your help!

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Your calculation seems good to me for the most part. You also have to show that it holds from below (that is, the limit as $x\to0$ for negative $x$). Approach it exactly the same and you'll get the same answer. – Clayton Feb 3 '13 at 20:11
@Clayton Good point, I didn't think of that. – Liam Feb 3 '13 at 20:16
No problem. It's a common mistake to make, and now you know for next time to be aware. – Clayton Feb 3 '13 at 20:21
Or you can just observe that $A(x)=x-2$ and then do an $\epsilon,\delta$ proof if you wish... – 1015 Feb 3 '13 at 21:05

In your proof you write "for $x=\frac12\epsilon$..." You need to prove the following holds $\forall x\in \mathbb{R}^*$ and not just for $x=\frac12\epsilon$ (it doesn't even cover the $x\to 0^-$ case)

Here is how I would do this. By additivity, $$A(x)=\int_{-2}^0f(t)\, dt+\int_0^xf(t)\, dt=\int_{-2}^0-1\, dt+\int_0^xf(t)\, dt=-2+\int_0^xf(t)\, dt$$ $A(0)=-2$. Now if $x>0$, $$A(x)=-2+\int_0^xf(t)\, dt=-2+\int_0^x1\, dt=-2+\left|x\right|$$ and if $x<0$, $$A(x)=-2+\int_0^xf(t)\, dt=-2+\int_0^x-1\, dt=-2-x=-2+\left|x\right|$$ So for $\epsilon>0$ and $\delta=\epsilon$, $$\left|x\right|<\delta\implies \left|A(x)-A(0)\right|=\left|-2+\left|x\right|+2\right|=\left|x\right|<\delta=\epsilon$$ and we are done.

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Yeah I added the $x \rightarrow 0^{-}$ case. It seems to me that if I prove for the cases $x=\frac{1}{2}\epsilon$ and $x=-\frac{1}{2}\epsilon$, then it does prove it for all real $x$ because $\epsilon$ could be any number in $\mathbb{R}^{+}$ and therefore $x$ could be any number in $\mathbb{R}^{*}$ – Liam Feb 3 '13 at 20:53
Actually, looking over it again, setting $x=\frac{1}{2}\epsilon$ was pretty superfluous, I could replace $\frac{1}{2}\epsilon$ with $x$ and use the fact that $\left|x\right|<\epsilon$, which I suppose is what you did – Liam Feb 3 '13 at 21:11
@Liam It is "wrong" to write $x=\frac 12 \epsilon$ and finish the proof with that $x$ – Nameless Feb 4 '13 at 12:52

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