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

  • #1
274
1

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)=[itex]x^{4}+\frac{1}{x}[/itex], a=1.


Homework Equations


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


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?
 

Answers and Replies

  • #2
699
5

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)=[itex]x^{4}+\frac{1}{x}[/itex], a=1.


Homework Equations


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


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 [itex]|x-1| < f(x)\epsilon[/itex]? Where f(x) is some expression.
$$
0<x-1<\delta
$$
 
  • #3
274
1
$$
|f(x) - L| = \left|x^4+\frac{1}{x} - 2\right| < \epsilon
$$
How can you make the above expression look like [itex]|x-1| < f(x)\epsilon[/itex]? 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 [itex]\delta_{1}=1[/itex]. Then [itex]|x-1|<1\Rightarrow-1<x-1<1\Rightarrow0<x<1[/itex]. Then [itex]0<x^{4}<16[/itex]. We then also have [itex]1/2<\frac{1}{x}[/itex].

That feels like it is a dead end not to mention that it doesn't look like itll become |x-1| anytime soon.
 
  • #4
Zondrina
Homework Helper
2,065
136
You need to use this definition :

[itex]\forallε>0, \existsδ>0 | 0 < |x-a| < δ \Rightarrow |f(x) - L| < ε[/itex]

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.

[itex]|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|}[/itex]

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
274
1
You need to use this definition :

[itex]\forallε>0, \existsδ>0 | 0 < |x-a| < δ \Rightarrow |f(x) - L| < ε[/itex]

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.

[itex]|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|}[/itex]

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...
 
  • #6
Zondrina
Homework Helper
2,065
136
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.
 
  • #7
274
1
Looks like its time to brush up on long division.... Thanks a lot! I think I can take it from here.
 
  • #8
274
1
Ok. Here is my attempt at the proof.

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

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

Then it follows that
[itex]
\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*}
[/itex]
This completes the proof.

How does this look?
 

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