Uniform Convergence of fn: Does fn(x)=nx^2/1+nx Converge?

[math8]

does {fn} converge uniformly? fn(x)=nx^2/1+nx


I can see that fn converges pointwise to f(x)=x. I know, for epsilon>0, I need to find N st for n >or equal to N, |fn(x)-f(x)|<epsilon.

|fn(x)-f(x)|=x/1+nx but then I am stuck.

 

[dirk_mec1]

does {fn} converge uniformly? fn(x)=nx^2/1+nx


I can see that fn converges pointwise to f(x)=x.

Why is that the case?

 

[statdad]

To be clear, are you considering convergence on all of $$\mathbb{R}$$ or
only on $$[0,1]$$? Also, your sequence of functions is

$$f_n(x) = \frac{x}{1+nx}$$

correct? If it really is (as you wrote)

$$f_n(x) = nx^2 + nx$$

then it should be clear that the only place it converges is at $$x = 0$$. Assuming the first version is correct, continue reading.

One idea: note that

$$|f_n(x) - x| = \frac{x}{1+nx} \le \frac{x}{nx} = \frac 1 n \quad \forall x$$

Given an $$\epsilon > 0$$, how would you choose an appropriate value of $$N$$?

 

[math8]

for dirk_mec1,

fn(x) converges pointwise to x because lim as n-->infinity of fn(x) equals x.

 

[dirk_mec1]

for dirk_mec1,

fn(x) converges pointwise to x because lim as n-->infinity of fn(x) equals x.

On what domain are these functions fn defined (like statdad already asked you) ?

 

[boombaby]

I will prefer the domain to be [0,infinity]...for x is not defined at -1/n, which makes it a bit complicated.