# Math Help - Proving a simple limit?

1. ## Proving a simple limit?

The following problem seems very simple, and indeed I'm sure the answer isn't hard at all. I know how to prove the limits of polynomial functions, using the epsilon-delta definition of a limit, by factoring certain parts, using certain theorems about sums and products of limits, and by finding upper bounds for certain expressions. From my understanding, the basic "approach" to finding the following:

$\lim_{x \to a} f(x)$

is to find a factor that puts the function in the following form:

$f(x) = (x-a)g(x) + b$

for some numbers $a$ and $b$, and some function $g(x)$. My problem is that I don't know how to go about doing so for the following problem:

Prove the following using the epsilon-delta definition of a limit:

$\lim_{h \to 0} f(a+h) = f(a)$

Now, obviously this limit is "intuitively" true but I can't seem to figure out the exact definition of delta in the proof. I've done some scratch work, I know that generally this forum encourages that posters display any work they've done themselves so far, so I'll put that below. Granted, it may contain errors:

I started with this:

$|h-0|=|h| < \delta$

and made the stipulation (with no particular motivation other then finding an upper bound on something):

$|h|< 1$

therefore:

$-1

the next line is something that I'm not certain is "always" true. In other words, I'm not certain if it follows from the above line:

$f(a-1)

now, I know I need to somehow get to the point of having this statement:

$|f(a+h) - f(a)|< \epsilon$

but I'm not sure how to proceed correctly. Do I subtract $f(a)$ from all parts of the inequality (the one three lines above) above? Or should I start by working firstly with $|f(a+h) - f(a)| < \epsilon \;\;$?

Any guidence would be much appreciated.

2. There have to be some restrictions on the function f(x) in order for this theorem to be true. For example, the theorem does not hold for the unit step function. Are you working only in polynomials? Only in continuous functions?

3. I made a mistake, the proper form of the question is this:

Prove the following:

$lim_{x \to a} f(x) = \lim_{h \to 0} f(a+h)$

given that the limit exists. Excuse my lack of clarity. I see that it would not be true, had I not supplied the full question as above.

4. For notation purposes suppose that $\lim _{x \to a} f(x) = L$.

Using the definition of limit: $\left( {\forall \varepsilon > 0} \right)\left( {\exists \delta > 0} \right)\left[ {0 < \left| {x - a} \right| < \delta \Rightarrow \left| {f(x) - L} \right| < \varepsilon } \right]$

Therefore, $\delta > \left| h \right| = \left| {a + h - a} \right| \Rightarrow \left| {f(a + h) - L} \right| < \varepsilon$.

5. Originally Posted by Plato
For notation purposes suppose that $\lim _{x \to a} f(x) = L$.

Using the definition of limit: $\left( {\forall \varepsilon > 0} \right)\left( {\exists \delta > 0} \right)\left[ {0 < \left| {x - a} \right| < \delta \Rightarrow \left| {f(x) - L} \right| < \varepsilon } \right]$

Therefore, $\delta > \left| h \right| = \left| {a + h - a} \right| \Rightarrow \left| {f(a + h) - L} \right| < \varepsilon$.
I'm still alittle lost as to how the line $|a + h - a| < \delta$ implies that $|f(a+h) - L| < \epsilon$ ? I'm sure it is an obvious implication, and that I'm just not seeing it; I just need alittle explanation of how "this" implies "that" in the above and I think I'll be able to understand the proof. I'm not questioning the validity of your proof, but rather I'm attempting to ensure my understanding of it.

6. Using the definition of limit: $\left( {\forall \varepsilon > 0} \right)\left( {\exists \delta > 0} \right)\left[ {0 < \left| {x - a} \right| < \delta \Rightarrow \left| {f(x) - L} \right| < \varepsilon } \right]$

${0 < \left| {\underbrace {(h + a)}_x - a} \right| < \delta \Rightarrow \left| {f(\underbrace {h + a}_x) - L} \right| < \varepsilon }$

7. Thank you, I see it bright and clear now.