Uniform Convergence of fn(x) in [0,1]?

[math8]

Let fn(x)=nx^2/1+nx ; x lies in [0,1]

Is the convergence uniform?

Since lim as n-->infinity of fn is x, I can see that fn(x) converges pointwise to f(x)= x
But I get stuck when I try to show the convergence is uniform or not.

 

[Dick]

Since you know the limit is f(x)=x, look at |fn(x)-f(x)|. What's the maximum of that function on the interval [0,1]?

 

[math8]

I think the maximum of |fn(x)-f(x)| on [0,1] is 1/1+n. So can I conclude that the convergence is uniform because there is an "N" that doesn't depend on x such that for all n> or eq. to N, 1/1+n gets arbitrarily small?

 

[Dick]

I think the maximum of |fn(x)-f(x)| on [0,1] is 1/1+n. So can I conclude that the convergence is uniform because there is an "N" that doesn't depend on x such that for all n> or eq. to N, 1/1+n gets arbitrarily small?

Can you make that conclusion? If I give you an e>0 can you find an N such that 1/(1+n)<e for all n>N? Sure you can. If you're not sure you'd better figure out how pick a corresponding N.

 

[math8]

that N should be > than 1-e/e right?

 

[Dick]

that N should be > than 1-e/e right?

Sure. Use parentheses when you write something like (1-e)/e, ok? 1-e/e at first glance looks like 1-(e/e), which looks like 0.

 

[math8]

Right, thanks.