Consider the sequence{fn} of functions fn:R→R defined by fn(x)=nx√1+n2x2 Find the pointwise limit of this sequence as n→∞. Does the sequence converge uniformly on R? Justify it.
Consider the sequence{fn} of functions fn:R→R defined by fn(x)=nx√1+n2x2
Find the pointwise limit of this sequence as n→∞.
Does the sequence converge uniformly on R? Justify it.
Solution:
To find the pointwise limit as n approaches infinity for the sequence of functions fn(x)=nx√1+n2x2 defined on the real numbers R, we'll examine the behavior of the function for each x in R as n tends to infinity.
As n approaches infinity, consider the function values for different values of x:
1. For x=0:
fn(0)=n.0√1+n2.0
Irrespective of the value of n,fn(0)=0 for all n, indicating the pointwise limit at x = 0 is 0.
2. For x ≠0:
fn(x)=nx√1+n2x2=x1n√1+n2x2
As n approaches infinity, the denominator 1n√1+n2x2 approaches ∣x∣, irrespective of the value of x. Thus, fn(x) tends towards x|x|=sgn(x), where sgn(x) is the signum function that yields -1 for negative x, 0 for x=0, and 1 for positive x.
Therefore, the pointwise limit of the sequence of functions fn(x) as n tends to infinity is a function that is 0 at x=0 and sgn(x) for x ≠0.
Now, to determine if the sequence converges uniformly on R, we must check whether the pointwise limit is also the limiting function for fn(x) uniformly.
For uniform convergence, we would need to show that the supremum of the difference between fn(x). and the limiting function tends to zero as n approaches infinity.
However, as the limiting function for x ≠0 is different from the limiting function at x=0 (0), the sequence of functions does not converge uniformly on R.
The pointwise limit is sgn(x) for x ≠0 and 0 at x=0, but this limit function does not represent the uniform convergence of the sequence.
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