Is Function f Continuous at x=0?

[IntroAnalysis]

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 x_o if x_oan element of domain and
lf(x) - f(x_o)l < ε whenever lx - x_ol < 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.

 

[SammyS]

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 x_o if x_oan element of domain and
lf(x) - f(x_o)l < ε whenever lx - x_ol < 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 $\displaystyle \lim_{x\to x_0}\,f(x)=f(x_0)\,,$ then f is continuous at x₀ .​

 

[IntroAnalysis]

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!

 

[SammyS]

What is the following limit?

$\displaystyle \lim_{x\to0}\,\frac{1}{(1+x)^{1/2} + 1}$​

 

[IntroAnalysis]

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.

 

[SammyS]

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!