Finding Delta to Satisfy |f(x)-l|<ε for 0<|x-a|<δ

[caduceus]

Homework Statement

Find a delta such that |f(x)-l|< epsilon for all x satisfying 0<|x-a|< delta

f(x)=(x^4)+(1/x); a=1; l=2

Homework Equations

Lemma Theorem

The Attempt at a Solution

0<|x-1|<delta; |(x^4)+(1/x)-2|<epsilon

|x^4-1|<epsilon/2 and |(1/x)-1|<epsilon/2

Then I applied third item of the Lemma which is

if y0 not equal to 0;

|y-y0|<min(|y0|/2,(epsilon*|y0|^2)/2)

then y not equal to 0 and;

|(1/y)-(1/y0)|<epsilon



Then I concluded that the |(1/x)-1|<epsilon/2 part of the solution as;

|x-1|<min(1/2,epsilon/2)

However, I could not find any approach for the |x^4-1|<epsilon/2

Which method shall I apply here?

Any helps will be appreciated.

 

[mighty2000]

Hello!

To find a suitable delta for this problem, we can use the definition of a limit.

First, we know that the limit of f(x) as x approaches a is l. In this case, a=1 and l=2.

So, we can start by writing the definition of a limit:

For any epsilon>0, there exists a delta>0 such that for all x satisfying 0<|x-a|<delta, we have |f(x)-l|<epsilon.

Now, let's substitute in the given values for a and l:

For any epsilon>0, there exists a delta>0 such that for all x satisfying 0<|x-1|<delta, we have |(x^4)+(1/x)-2|<epsilon.

We can simplify this expression by using the triangle inequality:

|(x^4)+(1/x)-2|<=|x^4-1|+|(1/x)-1|.

Now, we can focus on finding a suitable delta for each of these terms separately.

For the first term, |x^4-1|, we can use the fact that x^4-1=(x^2-1)(x^2+1). So, for any x satisfying 0<|x-1|<1, we can write:

|x^4-1|=|x^2-1||x^2+1|.

We can simplify further by using the fact that for any x satisfying 0<|x-1|<1, we have |x^2-1|<2|x-1|.

So, we can write:

|x^4-1|<2|x-1||x^2+1|.

Now, we can use the triangle inequality again to simplify this expression further:

|x^4-1|<2|x-1||x^2+1|<=2|x-1|(x^2+1).

For the second term, |(1/x)-1|, we can use the fact that for any x satisfying 0<|x-1|<1, we have |(1/x)-1|<2|x-1|.

So, now we have:

|(x^4)+(1/x)-2|<=|x^4-1|+|(1/x)-1|<2