Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
add comment

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.