# Thread: Definition of a Limit question.

1. ## Definition of a Limit question.

When proving that

$\lim_{x\rightarrow p}(ax+b)=ap+b,a>0$

using the definition of the limit (or the -separated definition), find the largest possible value of when a=2 and =001

2. $\forall \epsilon > 0, \ \exists \delta > 0$ such that whenever $0 < |x - p| < \delta$ it follows that $|(ax + b) - (ap + b)| < \epsilon$.

Looking at the latter expression and noting that a > 0:
$|(ax + b) - (ap + b)| = |ax - ap| = |a(x-p)| = a|x - p| < \epsilon$

So what $\delta$ should we choose?

3. hmm, im lost

4. What don't you understand?

Here's the formal definition of a limit:
$\forall \epsilon > 0, \ \exists \delta > 0$ such that whenever $0 < |x - p| < \delta$ it follows that $|f(x) - L| < \epsilon$

Here: $f(x) = ax + b$ and $L = ap + b$.

With these epsilon-delta proofs, we play around with the expression $|f(x)-L| < \epsilon$ to get $|x-p| < m$ for some $m$.

Why do we do this? Because we know that $|x - p| < \delta$. So if we let $\delta = a$, then through all the algebra we did, it must necessarily follow that $|f(x) - L| < \epsilon$.

By definition, this proves that $\lim_{x \to p} f(x) = L$

5. But how would go about actually solving for $\delta$, ie the largest value when $a=2$ and $\epsilon=0.01$

I guess I don't see where the answer lies in your explanation

6. What I gave you was the way to get the solution to $\delta$ in terms of both $\epsilon$ and $a$. Just plug it in and it should work.