Please check my proof
Let$\delta ,\epsilon >0$
$$|x-2|<\delta \rightarrow |f(x)-f(2)<\epsilon $$
$$ \rightarrow |3x^{2}-2x+1-3x^{2}-2x+1|<\epsilon $$
$$ \rightarrow |-4x|<\epsilon $$
$$ \rightarrow 4|-x|<\epsilon $$
$$ \rightarrow |-x|<\frac{\epsilon }{4}$$
choose $\delta =\frac{\epsilon }{4}$
then$ 4|-x|<\frac{4\epsilon }{4}=\epsilon $
therefore it 's continuous at 2
$\endgroup$ 42 Answers
$\begingroup$No, this is incorrect. There is a general "formula" for how these proofs should go. Whatever "$\delta$" calculations you do are in some way inessential. You should try again like so:
Let $\epsilon>0$. Suppose that $\delta=a>0$. Then $|x-2|<\delta$ implies that $|f(x)-f(2)|<\epsilon$. This asserts the existence of an appropriate $\delta$ for each $\epsilon$, which is what you need for continuity.
The "implies that" is the majority of the proof. You should plug in the function $f$ for both values, and show how your assumption on the size of $\delta$ can be used to prove the claim.
Additionally, you have made algebra errors in the "solution" for $\delta$. perhaps somebody will give a full solution, but as written, your proof is incorrect.
$\endgroup$ 4 $\begingroup$If $f(x)=3x^{2}-2x+1$, then $f(2)=3\cdot2^2-2\cdot2+1=9$. So you want to make $$\left|f(x)-f(2)\right|=\left|3x^{2}-2x+1-9\right|=\left|3x^{2}-2x-8\right|=\left|x-2\right|\left|3x+4\right|$$ arbitrarily small given that you can make $|x-2|$ sufficiently small. Can you proceed?
Two typical ways to proceed to find a suitable $\delta$ as a function of $\varepsilon$:
- first bound $\delta$ (e.g. $\delta \le 1$) to get an upper bound on the factor $\left|3x+4\right|$;
- rewrite: $3x+4=3(x-2)+10$, then $\left|3x+4\right| \le 3\left|x-2\right|+10$.
