Proving Limit of Function for -1 < x < 1, x≠0: Prove & Find Lim

In summary: The method of not using L'hopital's rule is just to use the first few terms of the Taylor series and show that it converges to the correct value. (I'm assuming you can find the Taylor series for that function--it's not hard.)For (ii), that's just saying that the derivative of the antiderivative is equal to the function itself. So if you take the derivative of the antiderivative, you should get back f(x).For (iii), they're just asking you to find constants that make the equation true. Think about how you would solve for those constants if you were given a polynomial and asked to find the coefficients. It's the same idea here, but
  • #1
tomboi03
77
0
For -1 < x < 1, x[tex]\neq[/tex]0 let
f(x)= [tex]\frac{\sqrt{1+x}-1}{x}[/tex]

i. Prove that
lim f(x)
x[tex]\rightarrow[/tex]0
exists and find it
(There is an easy proof and you get no credit for applying "hospital's rule")
Part (i) shows that f can be continued to a continuous function on (-1,1) if we assign this limit to be f(0) (this is assumed in subseguent parts)

ii. Show that the limit
lim 1/x [tex]\int[/tex] f(t) dt
x[tex]\rightarrow[/tex]0
exists

iii. Determine constants a0, a1, a2, a3 so that
[tex]\int[/tex] f(t) dt = a0 + a1x + a2x2 + a3x3 + x3 [tex]\rho[/tex](x)
(this integral goes from... 0 to x i couldn't figure out a way to put it on the integral.)
where
lim [tex]\rho[/tex](x) =0
x[tex]\rightarrow[/tex]0


I don't know how to go about this at all.

Please help me out
Thanks
 
Physics news on Phys.org
  • #2
Start with part (i). What must be true of [tex]\lim_{x\to0^{+}} f(x)[/tex] and [tex]\lim_{x\to0^{-}} f(x)[/tex] for [tex]\lim_{x\to0} f(x)[/tex] to exist?
 
  • #3
if both the integral approaches the same constant from the negative and positive side, then the function will approach that constant.

but, where do i go from there?
 
  • #4
tomboi03 said:
if both the integral approaches the same constant from the negative and positive side, then the function will approach that constant.

but, where do i go from there?

Integral; what integral? :confused: Don't you mean 'limit'?

For the limit to exist, both one-sided limits must approach the same value.

That means that if both one-sided limits approach the same value, then the limit exists.

So...do both one-sided limits approach the same value?
 
Last edited:
  • #5
For (i) you might want to rationalize the numerator.

For (ii) remember that an integral is always differentiable. What is the "difference quotient" for this?

For (iii) I don't understand what is wanted if f itself is not given explicitely.
 
  • #6
HallsofIvy said:
For (i) you might want to rationalize the numerator.

For (ii) remember that an integral is always differentiable. What is the "difference quotient" for this?

For (iii) I don't understand what is wanted if f itself is not given explicitely.

I assume f means the function of (i) (continued continuously at zero)
 
  • #7
Oh, yes, I see now. i, ii, and iii all come after the definition of f. But no integration is necessary for ii.
 
  • #8
tomboi03, think about a Taylor's series expansion.
 

1. What is the definition of a limit in calculus?

A limit in calculus is a fundamental concept that represents the behavior of a function as its input (x) approaches a certain value (a). It is denoted as lim x→a f(x), and it can be thought of as the value that a function approaches as its input gets closer and closer to the specified value.

2. How do you prove a limit of a function for a specific value?

In order to prove a limit of a function for a specific value, you need to show that the function approaches a single value as the input approaches the specified value. This can be done through various methods, such as algebraic manipulation, using the squeeze theorem, or using a graphing calculator to visually demonstrate the behavior of the function.

3. How do you find the limit of a function for a specific value?

To find the limit of a function for a specific value, you can either use direct substitution if the function is continuous at that value, or you can use algebraic manipulation or other methods to simplify the function and evaluate the limit. Additionally, you can also use the definition of a limit to find the limit of a function for a specific value.

4. Why is it important to prove and find limits of functions?

Proving and finding limits of functions is important in calculus because it helps us understand the behavior of a function as its input approaches a certain value. This knowledge is essential in many applications, such as optimization problems, finding derivatives, and determining the convergence of series.

5. What is the limit of a function for -1 < x < 1, x≠0?

The limit of a function for -1 < x < 1, x≠0 would be the same as the limit at x=0, as the limit only depends on the behavior of the function as it approaches the specified value. Therefore, the limit of a function for -1 < x < 1, x≠0 would be the same as the limit of the function at x=0.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
872
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
885
  • Calculus and Beyond Homework Help
Replies
12
Views
780
  • Calculus and Beyond Homework Help
Replies
2
Views
986
  • Calculus and Beyond Homework Help
Replies
4
Views
306
  • Calculus and Beyond Homework Help
Replies
2
Views
841
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
825
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
Back
Top