Is Function f Continuous at x=0?

  • Thread starter Thread starter IntroAnalysis
  • Start date Start date
  • Tags Tags
    Continuous
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
IntroAnalysis
Messages
58
Reaction score
0

Homework Statement


Define f:[-1,∞]→ℝ as follows: f(0) = 1/2 and

f(x) =[(1 + x)^(1/2) - 1]/x , if x ≠ 0

Show that f is continuous at 0.


Homework Equations


Definition. f is continuous at xo if xoan element of domain and
lf(x) - f(xo)l < ε whenever lx - xol < δ


The Attempt at a Solution


Do some algebra come up with f(x) = 1/[(1+x)^(1/2) + 1]

I also know that between [-1,1], f(x)≤ 1
and x ≥ 1, f(x) ≤ 1

I know I need to somehow pull out an lxl from the absolute value, since I know lxl≤δ then
I can define δ in terms of ε and the function.
 
Physics news on Phys.org
IntroAnalysis said:

Homework Statement


Define f:[-1,∞]→ℝ as follows: f(0) = 1/2 and   That should be f : [-1,∞) → ℝ

f(x) =[(1 + x)^(1/2) - 1]/x , if x ≠ 0

Show that f is continuous at 0.


Homework Equations


Definition. f is continuous at xo if xoan element of domain and
lf(x) - f(xo)l < ε whenever lx - xol < δ


The Attempt at a Solution


Do some algebra come up with f(x) = 1/[(1+x)^(1/2) + 1]

I also know that between [-1,1], f(x)≤ 1
and x ≥ 1, f(x) ≤ 1

I know I need to somehow pull out an lxl from the absolute value, since I know lxl≤δ then
I can define δ in terms of ε and the function.
Do you need to use an ε - δ proof, or can you use the limit criterion for continuity?
If [itex]\displaystyle \lim_{x\to x_0}\,f(x)=f(x_0)\,,[/itex] then f is continuous at x0 .​
 
It just says to show, it doesn't specify δ-ε proof. I've spent hours working on this one problem, any suggestions greatly appreciated!
 
It is 1/2 which is f(0).
So this approach I show: 1) the point c is in the domain
2) the limit of f(c) exists and
3) lim x->c f(x)=f(c)

I should have thought of this, it is a lot easier to show. Thank you.
 
IntroAnalysis said:
It is 1/2 which is f(0).
So this approach I show: 1) the point c is in the domain
2) the limit of f(c) exists and   more precisely, lim x→c f(x) exists
3) lim x->c f(x)=f(c)

I should have thought of this, it is a lot easier to show. Thank you.
You're welcome!