Find $\sup \delta$ such that $f(x)$ is $\epsilon-$close to $L$ Let $f(x) = x^3 + 3x^2 - x +1$. Let $c=2$. Fine the least upper bound of $\delta$ so that $f(x)$ is bounded within $\epsilon$ of $f(2)$ for all sufficiently small $\epsilon&gt;0$.


The solution says that


  In general, the bound is given by $\frac{1}{f'(c)}$ when $f$ is differentiable and $f'(c)\ne 0$.


Why $\frac{1}{f'(c)}$?
The derivative tells how fast a function in increasing at a point.

Intuitively (but not completely rigorously), if $|x-c| &lt; \delta$ for small enough $\delta$,

$$|f'(c)| \approx \frac{|f(x) - f(c)|}{|x-c|},$$

which implies

$$|f(x) - f(c)| \approx |x-c| \cdot |f'(c)| &lt; \delta |f'(c)|.$$

So in general, $\delta = \frac{1}{|f'(c)|}\epsilon$ will be your least upper bound of $\delta$.
The derivative tells how fast a function in increasing at a point.

Intuitively (but not completely rigorously), if $|x-c| &lt; \delta$ for small enough $\delta$,

$$|f'(c)| \approx \frac{|f(x) - f(c)|}{|x-c|},$$

which implies

$$|f(x) - f(c)| \approx |x-c| \cdot |f'(c)| &lt; \delta |f'(c)|.$$

So in general, $\delta = \frac{1}{|f'(c)|}\epsilon$ will be your least upper bound of $\delta$.