# Definition of Limit

1. Aug 23, 2008

### tanzl

1. The problem statement, all variables and given/known data

1)Prove that > limit(f(x), x = a) = limit(f(a+h), h = 0)

2. Prove that limit(f(x), x = a) = L iff limit(f(x)-L, x = a) = 0

2. Relevant equations

3. The attempt at a solution
1)I have tried to use the definition |f(x)-limf(a+h)| and |f(a+h)-limf(x)|. But it doesnt seem to be working because no further simplification can be done. I cant find a way to relate it back to |x-2|<delta and |h|<delta because the i cant eliminate the limit of function. I have also tried to assume both limit equals to f(a) and proves them. But again, it stucks.

2)First I assume that limit(f(x), x = a) = L. Then, the definition follows.
There exists 0<|x-a|<delta such that |f(x)-L|< epsilon.....(1)
So, in the case of limit(f(x)-L, x = a) = 0.
0<|x-a|<delta |f(x)-L-0|=|f(x)-L|<epsilon.... from (1)
and vice versa.
But it doesnt look like a proof to me. It more like I am rewriting it in another way. Is there a better way to put it?

Thanks...

2. Aug 23, 2008

### HallsofIvy

Staff Emeritus
"|f(x)- lim f(a+h)|" is very awkward. The DEFINITION of limit is that if $\lim_{x\rightarrow a}= L$, then, given any $\epsilon> 0$, there exist $\delta> 0$ such that if $0< |x- a|< \delta$ then $|f(x)-L|< \epsilon$. For $\lim_{h\rightarrow 0}f(x+ h)= L$, using that same definition, you need to show that given $\epsilon> 0$ there exist $\delta> 0$ such that if $0< |h- 0|= |h|<\delta$, then $|f(a+ h)- L|< \epsilon$. What happens if you replace "h" in the second formulation by x- a?

Rewriting it basically is the proof. The only difference between the first part and the second is that the function f(x) is replace by the function f(x)- L. Okay replace f(x) by that in the definition of limit:
Given $\epsilon> 0$ there exist $\delta$ such that if $0< |x- a|< \delta$ then $|(f(x)-L)- 0|= |f(x)- L|< \epsilon$. The fact that |(f(x)-L)- 0| is exactly the same as |f(x)- L| is the whole point.

3. Aug 23, 2008

### tanzl

For the 1st question, first we assume that
lim(f(x) , x = a) = L. Then, given any epsilon>0 there exists an delta>0 such that if
0<|x-a|<delta then |f(x)-L|<epsilon.

Then, the second part, we assume again lim(f(a+h) , h = 0) = L. Then, given any epsilon>0 there exists an delta>0 such that if
0<|h|<delta by letting |h|=|x-a| |f(a+h)-L|=|f(x)-L|<epsilon.

Is this the correct reasoning for the solution? It doesnt seem to be logical to me especially the assumption of lim(f(a+h) , h = 0) = L because if we do that we already assume that the both are equal so the prove is meaningless. But, if I assume lim(f(a+h) , h = 0) = M, it stucks.