Prove that $f(x) = 2x + 1$ is continuous in $p = 1$ using the $\delta - \epsilon$ definition

Question

I am learning the Delta - Epsilon definition, and got stuck in the first example. Maybe some of you can give me some light.

As in the title, I have to prove that $f(x) = 2x + 1$ is continuous in $p = 1$. So far I got that $|x| < \frac{\epsilon}{2}$, and then the book states that "Given $\epsilon > 0$ and taking $\delta = \frac{\epsilon}{2}$ we have: $1 - \delta < x < 1 + \delta \rightarrow f(1) - \epsilon < f(x) < f(1) + \epsilon$, then $f$ is continuous in $p = 1$."

I just don't understand why it takes $\delta = \frac{\epsilon}{2}$ and why the last statement proves $f$ is in fact continuous in $1$.

Any help is appreciated.

Answer 1

Your thoughts, I can see, are all over the place. Time to start over, and to move SLOWLY. The only way you will ever learn to do this if you do it very slowly, making sure to understand both why and how you are making each step.

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By definition, $f$ is continuous at $1$ if:

For every $\epsilon > 0$, there exists some $\delta>0$ such that if $|x-1|<\delta$, then $|f(x)-f(1)|<\epsilon$.

So, we are proving a statement that begins with "for every $\epsilon>0$. Therefore, our proof will start with the typical:

Let $\epsilon > 0$.

Now, we need to find some $\delta>0$ such that a condition will be met. That is, as long as we can choose one $\delta$, we are done. But which $\delta$ to pick? Well, the condition we need to meet is:

$$|f(x)-f(1)|<\epsilon$$

Well, write the left expression out and we get

$$|f(x)-f(1)| = |2x-1 - (2\cdot 1 - 1)| = |2x-1-2+1=2(x-1)|=2|x-1|$$

So, what we really need to meet is the condition:

$$2|x-1|<\epsilon$$

and at the same time, we can still choose a $\delta$ and demand that $|x-1|<\delta$. Well, we can see that if $|x-1|<\frac\epsilon2$, then $2|x-1|<\epsilon$.

Therefore, if we set $\delta=\frac\epsilon2$, then we know that if $|x-1|<\delta$, then $|f(x)-f(1)|<\epsilon$. OMG, we did it!

Answer 2

Note that for all $x\in\mathbb{R}$ and $y\in\mathbb{R}$ we have $$ |f(y)-f(x)|\leq 2\cdot |y-x| $$ Then given a $\epsilon>0$ we set $\delta=\epsilon/2$ and we have \begin{align} |y-x|<\delta=\frac{\epsilon}{2} \implies & 2\cdot |y-x|<2\cdot \delta={\epsilon} \\ \implies & |2y-2x|<2\cdot \delta={\epsilon} \\ \implies & |2y+1-2x-1|<2\cdot \delta={\epsilon} \\ \implies & |f(y)-f(x)|<2\cdot \delta={\epsilon} \\ \end{align}