Epsilon, Delta proofs of limits

[reaper_pball69]

Hey,
I'm having a little trouble with proofs of limits, I'm not sure but; it dosen't matter what value we choose for delta right or is that the epsilon value??

Anways can you help me with this example.

lim (x^3 - 3x^2 + 4x - 43)
x-2

Well i know the easy part L=-39 however im haing trouple with the proof; 0< l x-c l < delta and l f(x)-L l <epsilon.


Ya honours calculus is a killer.
Thanks guys.

 

[Matt]

Let f(x) = x³ - 3x² + 4x - 43.

Then |f(x) - L| = |f(x) + 39| = |(x - 2)(x² - x + 2)|.

 

[Euler]

I did this recently in my calculus AP class.

The way you find epsilon is

\(\displaystyle L-\epsilon<f(x)<L+\epsilon\)

Where f(x) is your original function and L is the limit, epsilon is your value between another y-value and the limit.

I usually just found delta with a table, heh.

 

[Matt]

I usually just found delta with a table, heh.

It is not possible to find delta in a numerical table when the epsilon is arbitrary.

 

[pka]

Usually I do not give such detail help. But, I confess to thinking at this is an type of question that has no place in beginning calculus.
If \(\displaystyle \
\varepsilon {\rm{ > 0}}
\\) then chose \(\displaystyle \
\delta {\rm{ = mim\{ 1,}}\varepsilon /14\}
\.\)
Looking at Matt’s first post: if \(\displaystyle \
\left| {x - 2} \right| < \delta
\\) then \(\displaystyle \|x| < 3\\) which implies \(\displaystyle \
\left| {x^2 - x + 2} \right| \le \left| {x^2 } \right| + \left| x \right| + 2 \le 9 + 3 + 2 = 14
\\)

Thus: \(\displaystyle \
\left| {x - 2} \right| < \delta \Rightarrow \left| {f(x) + 39} \right| = \left| {x - 2} \right|\left| {x^2 - x + 2} \right| < \left( {\frac{\varepsilon }{{14}}} \right)14 = \varepsilon .
\\)