GreenGoblin said:
Show that the following are continuous at x=1 using the epsilon-delta definition:
$x^{2} - x + 1$
$\sqrt (x)$
I know the definitions but I don't really know quite what to do with them. After the simple rearranging I'm just at a bit of a dead end; any pointers?
Gracias,
GreenGoblin
For a function to be continuous at a point, a limit needs to exist at that point.
By definition, if $\displaystyle 0 < |x - c| < \delta \implies \left|f(x) - L\right| < \epsilon$, then $\displaystyle \lim_{x \to c}f(x) = L$
So for the first one, to show that $\displaystyle \lim_{x \to 1}\left(x^2 - x + 1\right) = 1$, we need to show that $\displaystyle 0 < |x - 1| < \delta \implies \left|\left(x^2 - x + 1\right) - 1\right| < \epsilon$.
Trying to solve $\left|\left(x^2 - x + 1\right) - 1\right| < \epsilon $ for $\displaystyle |x - 1|$ gives us...
\[ \displaystyle \begin{align*} \left|x^2 - x\right| &< \epsilon \\ \left|x\left(x - 1\right)\right| &< \epsilon \\ |x||x - 1| &< \epsilon \end{align*} \]
Now define $\displaystyle M$ so that $\displaystyle |x| < M$ and then we have
\[ \displaystyle \begin{align*} M|x - 1| &< \epsilon \\ |x - 1| &< \frac{\epsilon}{M} \end{align*} \]
Suppose that we make $\displaystyle |x - 1| < \frac{1}{2}$, in other words, ensuring that the distance from x and 1 is never any more than 1/2 a unit (letting $\displaystyle \delta = \frac{1}{2}$, which we can do because we are going to close in on x = 1 by making that distance small anyway), and we find...
\[ \displaystyle \begin{align*} |x - 1| &< \frac{1}{2} \\ -\frac{1}{2} < x - 1 &< \frac{1}{2} \\ \frac{1}{2} < x &< \frac{3}{2} \end{align*} \]
and therefore we can let $\displaystyle M = \frac{3}{2}$, so
\[ \displaystyle \begin{align*} |x - 1| &< \frac{\epsilon}{\frac{3}{2}} \\ |x - 1| &< \frac{2}{3}\epsilon \end{align*} \]
Therefore, we can define $\displaystyle \delta = \min\left\{ \frac{1}{2}, \frac{2}{3}\epsilon \right\}$ and reverse each step, and you will have your proof :)