Consider the sequence`{f_n}` of functions `f_n:R \to R` defined by `f_n (x) = \frac{nx}{\sqrt {1 + n^2x^2} }` Find the pointwise limit of this sequence as `n \to \infty.` Does the sequence converge uniformly on R? Justify it.

Consider the sequence`{f_n}` of functions `f_n:R \to R` defined by `f_n (x) = \frac{nx}{\sqrt {1 + n^2x^2} }`

Find the pointwise limit of this sequence as `n \to \infty.`

Does the sequence converge uniformly on R? Justify it.

Solution:

To find the pointwise limit as n approaches infinity for the sequence of functions `f_n (x) = \frac{nx}{\sqrt {1 + n^2x^2} }` 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:

`f_n (0) = \frac{n.0}{\sqrt {1 + n^2 .0 }}`

Irrespective of the value of n,` f_n (0) = 0` for all n, indicating the pointwise limit at x = 0 is 0.

2. For x ≠ 0:

`f_n (x) = \frac{nx}{\sqrt {1 + n^2 x^2} } = \frac{x}{\frac{1}{n}\sqrt {1 + n^2 x^2} }`

 As n approaches infinity, the denominator `\frac{1}{n}\sqrt {1 + n^2 x^2}`  approaches ∣x∣, irrespective of the value of x. Thus, `f_n (x)` tends towards `\frac{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 `\f_n (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 `f_n (x)` uniformly.

For uniform convergence, we would need to show that the supremum of the difference between `f_n (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.