Need help with epsilon delta proof of f(x)=x^4+(1/x) as x goes to 1

1. Sep 10, 2012

1. The problem statement, all variables and given/known data
Determine the limit l for a given a and prove that it is the limit by showing how to find δ such that |f(x)-l|<ε for all x satisfying 0<|x-a|<δ.

f(x)=$x^{4}+\frac{1}{x}$, a=1.

2. Relevant equations
I claim that $\lim\limits_{x\rightarrow 1}x^{4}+\frac{1}{x}=2$.

3. The attempt at a solution
I tried things like solving for x but all my attempts were fruitless. Could anyone point me in the right direction?

2. Sep 10, 2012

Dustinsfl

$$|f(x) - L| = \left|x^4+\frac{1}{x} - 2\right| < \epsilon$$
How can you make the above expression look like $|x-1| < f(x)\epsilon$? Where f(x) is some expression.
$$0<x-1<\delta$$

3. Sep 10, 2012

I thought I could factor it but I wasn't able to.

Can I say, let $\delta_{1}=1$. Then $|x-1|<1\Rightarrow-1<x-1<1\Rightarrow0<x<1$. Then $0<x^{4}<16$. We then also have $1/2<\frac{1}{x}$.

That feels like it is a dead end not to mention that it doesn't look like itll become |x-1| anytime soon.

4. Sep 10, 2012

Zondrina

You need to use this definition :

$\forallε>0, \existsδ>0 | 0 < |x-a| < δ \Rightarrow |f(x) - L| < ε$

You want to find a delta that will satisfy this definition. The trick with these sorts of problems is to massage |f(x)-L| until you are in the form |x-a| < δ.

$|f(x) - L| = \left|x^4+\frac{1}{x} - 2\right| = |\frac{x^5-2x+1}{x}| = |x-1||x^4+x^3+x^2+x-1|\frac{1}{|x|}$

Now ask yourself... using the definition, what can you do to manipulate this expression into a suitable δ.

EDIT : Also the triangle inequality will be useful here : |a+b| ≤ |a| + |b|

5. Sep 10, 2012

Is there a certain way to factor out functions like that? I couldn't for the life of me figure out how to factor it...

6. Sep 10, 2012

Zondrina

I cheated a bit since i know that the |x-a| factor is ALWAYS going to show up when you massage your |f(x)-L| expression. I used polynomial long division coupled with the factor & remainder theorems.

7. Sep 10, 2012

Looks like its time to brush up on long division.... Thanks a lot! I think I can take it from here.

8. Sep 10, 2012

Ok. Here is my attempt at the proof.

Proof: Suppose we are given ε>0. Then choose δ=min(1,$\frac{2\epsilon}{31})$.
Thus we have
\begin{align*} 0<|x-1|<δ &\Rightarrow|x-a|<\frac{2\epsilon}{31}. \end{align*}

Now we will set $|x-1|<1\Rightarrow-1\leq x-1\leq 1\Rightarrow 0<x<2.$ Note that $|x-1|\leq|x|+|-1|=|x|+1$. Then since |x|<2, we have $|x|+1<2+1=3.$ Thus $|x-1|<3$/
Note that $|x^{4}+x^{3}+x^{2}+x-1|\leq|x^{4}+x^{3}+x^{2}|+|x-1|<|x^{4}+x^{3}+x^{2}|+3\leq|x^{4}+x^{3}|+|x^{2}|+3.$ Now since |x|<2, we have $|x^{2}|<2^{2}=4$ so $|x^{4}+x^{3}|+|x^{2}|+3<|x^{4}+x^{3}|+4+3=|x^{4}+x^{3}|+7\leq|x^{4}|+|x^{3}|+7$. Now $|x^{4}|<2^{4}=16$ and $|x^{3}|<2^{3}=8$ so $|x^{4}|+|x^{3}|+7<16+8+7=31$. Thus $|x^{4}+x^{3}+x^{2}+x-1|<31$.

Then it follows that
\begin{align*} |x-a|<\frac{2\epsilon}{31} &\Rightarrow\frac{1}{|x|}|x-1||x^{4}+x^{3}+x^{2}+x-1|<\frac{2\cdot 31\cdot\epsilon}{2\cdot 31}\\ &\Rightarrow \frac{1}{|x|}|x-1||x^{4}+x^{3}+x^{2}+x-1|<\epsilon\\ &\Rightarrow \frac{|x-1||x^{4}+x^{3}+x^{2}+x-1|}{|x|}<\epsilon\\ &\Rightarrow \frac{|x^{5}-2x+1|}{|x|}<\epsilon\\ &\Rightarrow |x^{4}+\frac{1}{x}-2|<\epsilon. \end{align*}
This completes the proof.

How does this look?