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

[DeadOriginal]

Homework Statement

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.

Homework Equations

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

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?

 

[Dustinsfl]

Homework Statement

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.

Homework Equations

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

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?

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

 

[DeadOriginal]

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

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

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

 

[STEMucator]

You need to use this definition :

\forallε&gt;0, \existsδ&gt;0 | 0 &lt; |x-a| &lt; δ \Rightarrow |f(x) - L| &lt; 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|

 

[DeadOriginal]

You need to use this definition :

\forallε&gt;0, \existsδ&gt;0 | 0 &lt; |x-a| &lt; δ \Rightarrow |f(x) - L| &lt; 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|

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...

 

[STEMucator]

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...

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.

 

[DeadOriginal]

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

 

[DeadOriginal]

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
<br /> \begin{align*}<br /> 0&lt;|x-1|&lt;δ &amp;\Rightarrow|x-a|&lt;\frac{2\epsilon}{31}.<br /> \end{align*}<br />

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

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

How does this look?