$\varepsilon$-$\delta$-proof: $f(x)=2x+1$ is continuous at $x_{0}=2$

Question

Via $\varepsilon$-$\delta$, prove that the function $f(x)=2x+1$ is continuous at $x_{0}=2$

Let $\varepsilon > 0$, let $\delta > 0$ and $|x - x_{0}| < \delta$

$\Rightarrow$

$$|2x+1 - (2x_{0}+1)| = |2x+1 - 2x_{0}-1| = |2x-2x_{0}| = |2(x-x_{0})| < |2\delta|=\varepsilon$$

$\Rightarrow$

$$2\delta = \varepsilon$$

$$\delta = \frac{\varepsilon}{2}$$

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I'm pretty sure this will be a part of our exam so please tell me if I did it correctly. I have the feeling there is something missing at the end but I'm really not sure.

Answer 1

In general, it is correct. If you want a very precise proof, you shall redact something like:

Let $\epsilon>0$ and set $\delta = \tfrac{\epsilon}{2}$. If $|x-x_{0}|<\delta$, then $$ |2x+1 - (2x_{0}+1)| = |2x+1 - 2x_{0}-1| = |2x-2x_{0}| = 2|x-x_{0}| < 2\delta=\varepsilon, $$ proving that $f(x)=2x+1$ is continuous at $x_{0}$.

Answer 2

You actually proved that $f(x)=2x+1$ is continuous for any $x_0$ which is way better (but definitely make sure you know the difference).

Fortunately this is a linear function so it isn't too hard. Just for the sake of future more difficult delta epsilon proofs it is worth noting that to prove this exact statement it is sufficient to show that $$ |x-2|<\delta\Rightarrow |2x-1-(2*2-1)|=|2x-4|<\epsilon $$ Which you can conclude quickly as the above.