1. ## Proving A Limit

Is this proof correct?

$\lim_{x\to0}(4x+7)=7$

Proof:
$\mbox{Choose } \delta=\frac{\epsilon}{4} \mbox{THEN IF} \
0<|x-0|<\frac{\epsilon}{4} \ \mbox{THEN} \
|f(x)-7| < \epsilon$

2. Originally Posted by Mike9182
Is this proof correct?

$\lim_{x\to0}(4x+7)=7$

Proof:
$\mbox{Choose } \delta=\frac{\epsilon}{4} \mbox{THEN IF} \
0<|x-0|<\frac{\epsilon}{4} \ \mbox{THEN} \
|f(x)-7| < \epsilon$
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):

$\forall\varepsilon>0,\,\exists\,\delta>0 : 0<\left|x-0\right|<\delta\implies\left|(4x+7)-7\right|<\varepsilon$.

However, $\left|(4x+7)-7\right|=\left|4x\right|=4\left|x\right|<\varepsil on\implies\left|x\right|<\frac{\varepsilon}{4}=\de lta$.

Thus, take $\delta = \frac{\varepsilon}{4}$. Therefore, $\lim_{x\to0}(4x+7)=7$

3. Originally Posted by Chris L T521
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):

$\forall\varepsilon>0,\,\exists\,\delta>0 : 0<\left|x-0\right|<\delta\implies\left|(4x+7)-7\right|<\varepsilon$.

However, $\left|(4x+7)-7\right|=\left|4x\right|=4\left|x\right|<\varepsil on\implies\left|x\right|<\frac{\varepsilon}{4}=\de lta$.

Thus, take $\delta = \frac{\varepsilon}{4}$. Therefore, $\lim_{x\to0}(4x+7)=7$
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?

4. Originally Posted by VonNemo19
What part of this actually proves the limit?
The proof lies in the fact that $\exists\,\delta:0<\left|x-c\right|<\delta\implies\left|f\!\left(x\right)-L\right|<\varepsilon$, and we found in the problem that $\delta=\frac{\varepsilon}{4}>0 \backepsilon \varepsilon>0\,(\forall\,\varepsilon)$

To find out more about $\varepsilon-\delta$ proofs, look at this thread: http://www.mathhelpforum.com/math-he...ta-proofs.html

5. Originally Posted by Chris L T521
The proof lies in the fact that $\exists\,\delta:0<\left|x-c\right|<\delta\implies\left|f\!\left(x\right)-L\right|<\varepsilon$, and we found in the problem that $\delta=\frac{\varepsilon}{4}>0 \backepsilon \varepsilon>0\,(\forall\,\varepsilon)$

To find out more about $\varepsilon-\delta$ proofs, look at this thread: http://www.mathhelpforum.com/math-he...ta-proofs.html
I get it. Thanks man.