# Thread: Epsilon - Delta Proof Question

1. ## Epsilon - Delta Proof Question

lim ((1/(x+4)) = 1/2
x-> -2

or

a=-2
L=1/2
f(x)= (1/(x+4))

The question says we will need a constraint on |x+2|, so it says to assume |x+2|<1.
This is the part that is confusing me because I can solve any other limit question.

a)[0,min{1,4epsilon})
b)(0,min{1,6epsilon}]
c)[0,min{1,2epsilon})
d)(0,min{1,2epsilon}]
e)(0,min{1,4epsilon})

2. ## Re: Epsilon - Delta Proof Question

Originally Posted by poipoipoi10
lim ((1/(x+4)) = 1/2
x-> -2
The question says we will need a constraint on |x+2|, so it says to assume |x+2|<1.
This is the part that is confusing me because I can solve any other limit question.
a)[0,min{1,4epsilon})
b)(0,min{1,6epsilon}]
c)[0,min{1,2epsilon})
d)(0,min{1,2epsilon}]
e)(0,min{1,4epsilon})
Notice that $\displaystyle \left| {\frac{1}{{x + 4}} - \frac{1}{2}} \right| = \frac{1}{2}\frac{{\left| {x + 2} \right|}}{{\left| {x + 4} \right|}}$.

If $\displaystyle |x+2|<1$ what can you say about $\displaystyle \frac{1}{|x+4|}~?$

3. ## Re: Epsilon - Delta Proof Question

Sigh... I'm confused -.-

I guess that
-1<|x+2|<1
-3<x<-1
1<x+4<3

sooooo uh yeah... haha

4. ## Re: Epsilon - Delta Proof Question

Originally Posted by poipoipoi10
Sigh... I'm confused -.-
Here a hint.
$\displaystyle |x+4|\ge 1$

5. ## Re: Epsilon - Delta Proof Question

So would the answer be D???
Because if you plug 1 as the |x+4|, from the equation you gave me: 1|x+2|/2|x+4|<epsilon
I would get |x+2|<2*epsilon
I don't really know if i've done that correctly, I've been doing work all day and my head is swimming with different subjects

6. ## Re: Epsilon - Delta Proof Question

Originally Posted by poipoipoi10
lim ((1/(x+4)) = 1/2
x-> -2

or

a=-2
L=1/2
f(x)= (1/(x+4))

The question says we will need a constraint on |x+2|, so it says to assume |x+2|<1.
This is the part that is confusing me because I can solve any other limit question.

a)[0,min{1,4epsilon})
b)(0,min{1,6epsilon}]
c)[0,min{1,2epsilon})
d)(0,min{1,2epsilon}]
e)(0,min{1,4epsilon})
To show $\displaystyle \displaystyle \lim_{x \to c}f(x) = L$ you need to prove $\displaystyle \displaystyle 0 < |x - c| < \delta \implies |f(x) - L| < \epsilon$. So you need to show $\displaystyle \displaystyle 0 < |x + 2| < \delta \implies \left|\frac{1}{x + 4} - \frac{1}{2}\right| < \epsilon$.

Scratch work:

\displaystyle \displaystyle \begin{align*} |f(x) - L| &< \epsilon \\ \left|\frac{1}{x + 4} - \frac{1}{2}\right| &< \epsilon \\ \left|-\left[\frac{x+2}{2(x + 4)}\right]\right| &< \epsilon \\ \frac{|x + 2|}{2|x + 4|} &< \epsilon \\ |x + 2| &< 2|x + 4|\epsilon \end{align*}

But we can't let $\displaystyle \displaystyle \delta = 2|x + 4|\epsilon$ because $\displaystyle \displaystyle \delta$ is only supposed to be a function of $\displaystyle \displaystyle \epsilon$, not a function of $\displaystyle \displaystyle x$. But if we remember that the act of taking a limit means that we are showing that moving $\displaystyle \displaystyle x$ closer to the value of $\displaystyle \displaystyle c$ in both directions moves $\displaystyle \displaystyle f(x)$ closer to the value of $\displaystyle \displaystyle L$ in both directions, that means we can choose a value of $\displaystyle \displaystyle x$ as close to $\displaystyle \displaystyle c$ as we like, then make it move closer. So say we want to start off with $\displaystyle \displaystyle x$ no more than $\displaystyle \displaystyle 1$ unit away from $\displaystyle \displaystyle c$, then $\displaystyle \displaystyle |x + 2| < 1$. Therefore

\displaystyle \displaystyle \begin{align*} |x + 2| &< 1 \\ -1 < x + 2 &< 1 \\ 2 < x + 4 &< 3 \\ |x + 4| &< 3 \\ 2|x + 4| &< 6 \\ 2|x + 4|\epsilon &< 6\epsilon \end{align*}

So we can let $\displaystyle \displaystyle \delta = \min\left\{1, 6\epsilon\right\}$ and reverse each step to get your proof.