# Thread: Another Epsilon Delta proof?

1. ## Another Epsilon Delta proof?

Heres another problem that I haven't been able to figure out, I need some assistance:

Suppose there is a $\displaystyle \delta > 0$ such that $\displaystyle f(x) = g(x)$ when:

$\displaystyle 0 < |x-a| < \delta$

Prove that:

$\displaystyle \lim_{x \to a} f(x) = \lim_{x \to a} g(x)$

2. Sounds like a very good teacher!

I once taught a course called "Quantitative Methods for Economics and Business Administration" using a text book assigned by the Business Administration department. On one page of this book we were given three laws of limits:
1) If $\displaystyle \lim_{x\to a} f(x)= L$ and $\displaystyle \lim_{x\to a} g(x)= M$ then $\displaystyle \lim_{x\to a} (f+g)(x)= L+ M$
2) If $\displaystyle \lim_{x\to a} f(x)= L$ and $\displaystyle \lim_{x\to a} g(x)= M$ then $\displaystyle \lim_{x\to a} fg(x)= LM$
3) if $\displaystyle \lim_{x\to a} f(x)= L$ and $\displaystyle \lim_{x\to a} g(x)= M$ and M is not 0, then $\displaystyle \lim_{x\to a} \frac{f(x)}{g(x)}= \frac{L}{M}$

On the very next page, they introduced the derivative as the limit of the difference quotient: $\displaystyle f'(a)= \lim_{h\to 0}\frac{f(a+h)- f(a)}{h}$, completely ignoring the fact that the denominator necessarily goes to 0 so this limit could NEVER be done using only those laws!

Yes, that property is very important, and fairly easy to prove.

Let $\displaystyle \lim_{x\to a} f(x)= L$. Then, by definition of limit, given any $\displaystyle \epsilon> 0$ there exist $\displaystyle \delta_1> 0$ such that if $\displaystyle 0<|x- a|< \delta_1$ then $\displaystyle |f(x)- L|< \epsilon$. Note the "<" symbol at the left of |x-a|. It does not matter what happens at x= a because then |x-a|= 0! (I am using the subscript on $\displaystyle \delta_1$ to distinguish if from the $\displaystyle \delta$ given in the problem.)

Now, given any $\displaystyle \epsilon> 0$, we can use the smaller of $\displaystyle \delta$ and $\displaystyle \delta_1$ to prove that $\displaystyle \lim_{x\to a} g(x)= L$. If 0< |x- a|< the smaller of $\displaystyle \delta$ and $\displaystyle \delta_1$, then both $\displaystyle |f(x)- L|< \epsilon$ and f(x)= g(x) are true. So we can replace f(x) by g(x) in that expression: $\displaystyle |g(x)- L|< \epsilon$ and we are done!

3. hint:

choose delta=epsilon

4. Originally Posted by HallsofIvy
I once taught a course called "Quantitative Methods for Economics and Business Administration" using a text book assigned by the Business Administration department. On one page of this book we were given three laws of limits:
1) If $\displaystyle \lim_{x\to a} f(x)= L$ and $\displaystyle \lim_{x\to a} g(x)= M$ then $\displaystyle \lim_{x\to a} (f+g)(x)= L+ M$
2) If $\displaystyle \lim_{x\to a} f(x)= L$ and $\displaystyle \lim_{x\to a} g(x)= M$ then $\displaystyle \lim_{x\to a} fg(x)= LM$
3) if $\displaystyle \lim_{x\to a} f(x)= L$ and $\displaystyle \lim_{x\to a} g(x)= M$ and M is not 0, then $\displaystyle \lim_{x\to a} \frac{f(x)}{g(x)}= \frac{L}{M}$

On the very next page, they introduced the derivative as the limit of the difference quotient: $\displaystyle f'(a)= \lim_{h\to 0}\frac{f(a+h)- f(a)}{h}$, completely ignoring the fact that the denominator necessarily goes to 0 so this limit could NEVER