Epsilon Delta Limit Definition

In summary: The expression isn't going to simplify very nicely. We need to put a bound on it.In summary, to prove the limit using epsilon, delta definition, we need to find a delta > 0 such that if 0 < |x + 1| < delta, then |[1/(sqrt((x^2)+1)]-(sqrt(2))/2| < epsilon. To find this delta, we can rationalize the numerator to get 2(1-x^2) in the numerator and then restrict delta to put a bound on the rest of the expression.
  • #1
MathSquareRoo
26
0

Homework Statement



Prove lim x--> -1

1/(sqrt((x^2)+1)

using epsilon, delta definition of a limit

Homework Equations





The Attempt at a Solution



I know that the limit =(sqrt(2))/2

And my proof is like this so far. Let epsilon >0 be given. We need to find delta>0 s.t. if 0<lx+1l<delta, then l[1/(sqrt((x^2)+1)]-(sqrt(2))/2l < epsilon. So we need to pick delta= ?

I'm not sure how to arrive at the delta. I know I have to work out what's inside the absolute values, but I'm getting stuck.
 
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  • #2
MathSquareRoo said:

Homework Statement



Prove lim x--> -1

1/(sqrt((x^2)+1)

using epsilon, delta definition of a limit

Homework Equations



The Attempt at a Solution



I know that the limit =(sqrt(2))/2

And my proof is like this so far. Let epsilon >0 be given. We need to find delta>0 s.t. if 0<lx+1l<delta, then l[1/(sqrt((x^2)+1)]-(sqrt(2))/2l < epsilon. So we need to pick delta= ?

I'm not sure how to arrive at the delta. I know I have to work out what's inside the absolute values, but I'm getting stuck.
What have you tried?

Where are you stuck?
 
  • #3
I don't know how to simplify it at all. My thought was to maybe to get common denominator or and then multiply by the conjugate, but I don't know if this is correct. I got l[2-(2)^(1/2) *((x^2)+1)^(1/2)]/(2((x^2)+1)^(1/2)l. I'm sorry, it's hard to type it here, does any of that make sense?
 
  • #4
MathSquareRoo said:
I don't know how to simplify it at all. My thought was to maybe to get common denominator or and then multiply by the conjugate, but I don't know if this is correct. I got l[2-(2)^(1/2) *((x^2)+1)^(1/2)]/(2((x^2)+1)^(1/2)l. I'm sorry, it's hard to type it here, does any of that make sense?
Yes, that makes some sense.

What you have is: [itex]\displaystyle \left|\frac{2-\sqrt{2}\sqrt{1+x^2}}{2\,\sqrt{1+x^2}}\right|[/itex] which is equivalent to: [itex]\displaystyle \left|\frac{1}{\sqrt{1+x^2}}-\frac{\sqrt{2}}{2}\right|[/itex]. That's a start.

Rationalize the numerator. One factor of the result will be (x + 1) .

Restrict δ to put a bound on the rest of the expression.
 
  • #5
When I try to rationalize, I get l(-2x^2)/[4((x^2)+1)^(1/2)+2(2)^(1/2)*((x^2)+1)]l

What did I do wrong?
 
  • #6
MathSquareRoo said:
When I try to rationalize, I get l(-2x^2)/[4((x^2)+1)^(1/2)+2(2)^(1/2)*((x^2)+1)]l

What did I do wrong?
The numerator should be 2(1-x2) which is 1(1-x)(1+x).

The denominator is a mess.
 

1. What is the Epsilon Delta Limit Definition?

The Epsilon Delta Limit Definition is a way of mathematically defining the limit of a function as a specific value is approached. It is used to prove the existence of a limit and to determine the value of the limit.

2. How does the Epsilon Delta Limit Definition work?

The Epsilon Delta Limit Definition works by setting a tolerance, or "epsilon" value, and a range, or "delta" value, around the limit point. If the function values stay within this range as the input values get closer to the limit point, then the limit exists and its value can be determined.

3. Why is the Epsilon Delta Limit Definition important?

The Epsilon Delta Limit Definition is important because it is a rigorous and precise way of determining the value of a limit. It also helps to prove the existence of a limit, which is crucial in many mathematical and scientific applications.

4. Can you give an example of using the Epsilon Delta Limit Definition?

Sure, let's say we have the function f(x) = 2x + 3 and we want to find the limit as x approaches 2. We can set an epsilon value of 0.1 and a delta value of 0.05. This means that if |x - 2| < 0.05, then |f(x) - 7| < 0.1. By choosing a small enough delta value, we can ensure that the function values will stay within the tolerance range, and thus the limit exists and its value is 7.

5. Are there any limitations to the Epsilon Delta Limit Definition?

Yes, there are some limitations to the Epsilon Delta Limit Definition. It can only be used for functions that approach a single limit point, and it may not work for more complex functions. Additionally, it can be a time-consuming process to determine the epsilon and delta values, and it may not always be practical to use in real-world scenarios.

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