Is this proof correct?
$\displaystyle \lim_{x\to0}(4x+7)=7$
Proof:
$\displaystyle \mbox{Choose } \delta=\frac{\epsilon}{4} \mbox{THEN IF} \
0<|x-0|<\frac{\epsilon}{4} \ \mbox{THEN} \
|f(x)-7| < \epsilon $
Yup, that is correct!
I would state it this way, though (its up to you, but I would show a little more work that that):
$\displaystyle \forall\varepsilon>0,\,\exists\,\delta>0 : 0<\left|x-0\right|<\delta\implies\left|(4x+7)-7\right|<\varepsilon$.
However, $\displaystyle \left|(4x+7)-7\right|=\left|4x\right|=4\left|x\right|<\varepsil on\implies\left|x\right|<\frac{\varepsilon}{4}=\de lta$.
Thus, take $\displaystyle \delta = \frac{\varepsilon}{4}$. Therefore, $\displaystyle \lim_{x\to0}(4x+7)=7$
What part of this actually proves the limit? I follow everything, but I am unsure about the implications of how the last line proves that the limit exists. Is this because it has been shown that given any value of epsilon, delta will be less than one quarter of epsilon?
The proof lies in the fact that $\displaystyle \exists\,\delta:0<\left|x-c\right|<\delta\implies\left|f\!\left(x\right)-L\right|<\varepsilon$, and we found in the problem that $\displaystyle \delta=\frac{\varepsilon}{4}>0 \backepsilon \varepsilon>0\,(\forall\,\varepsilon)$
To find out more about $\displaystyle \varepsilon-\delta$ proofs, look at this thread: http://www.mathhelpforum.com/math-he...ta-proofs.html