Proving Limits: Definition & Equations

[tanzl]

1. Homework Statement

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

Homework Equations

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 doesn't seem to be working because no further simplification can be done. I can't find a way to relate it back to |x-2|<delta and |h|<delta because the i can't 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 doesn't 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...

 

[HallsofIvy]

1. Homework Statement

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

Homework Equations

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 doesn't seem to be working because no further simplification can be done. I can't find a way to relate it back to |x-2|<delta and |h|<delta because the i can't eliminate the limit of function. I have also tried to assume both limit equals to f(a) and proves them. But again, it stucks.

"|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?

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 doesn't 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...

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.

 

[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 doesn't 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.