Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
|f(x) - L| < E whenever |x - c|< d:
lim (4x - 1) = 11
x -> 3
Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
|f(x) - L| < E whenever |x - c|< d:
lim (4x - 1) = 11
x -> 3
You want to find $\displaystyle \delta$ such that $\displaystyle |x-3|<\delta \implies |(4x-1)-11|<\epsilon$.
Choose $\displaystyle \delta=\frac{\epsilon}{4}$.
Thus when $\displaystyle |x-3|<\delta$, we have that $\displaystyle |(4x-1)-11|=|4x-12|=4|x-3|<4\delta=\frac{4\epsilon}{4}=\epsilon$
If you don't know what $\displaystyle \delta$ is, you can solve for it.
Whatever $\displaystyle \delta$ is you know that $\displaystyle |x-3|<\delta$.
So, work with what you have. You ultimately want $\displaystyle |(4x-1)-11|<\epsilon$.
So, similar to before,
$\displaystyle |(4x-1)-11|=|4x-12|=4|x-3|<4\delta$.
Now you want $\displaystyle 4\delta$ to be equal to $\displaystyle \epsilon$, so you just solve for it.
$\displaystyle 4\delta=\epsilon\implies \delta=\frac{\epsilon}{4}$