# Thread: Definition of a Limit question.

1. ## Definition of a Limit question.

When proving that

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

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

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

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

So what $\displaystyle \delta$ should we choose?

3. hmm, im lost

4. What don't you understand?

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

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

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

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

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

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

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 $\displaystyle \delta$ in terms of both $\displaystyle \epsilon$ and $\displaystyle a$. Just plug it in and it should work.