On uniform convergence of sequence

Suppose (f_n} is a sequence of functions where f_n(x) = x / (1 + n^2 x^2).

I am finding the pointwise limit of the sequence of {f_n'(x)} on the interval
(-oo, + oo)...in which {f_n'(x)} is the sequence of functions obtained from the derivative of x / (1 + n^2 x^2) and I am trying to find out if this {f_n'(x)}
converges uniformly or not.

SOlution:

Well, the derivative of x / (1 + n^2 x^2) is :

(1 - n^2 x^2) / ( 1 + n^2 x^2 )^2.

Now,
when x = 0, lim (n -> +oo) f_n'(0) = (1 - n^2 0^2) / ( 1 + n^2 0^2 )^2 = 1

When x is not 0,

lim (n -> +oo) f_n'(x) = (1 - n^2 x^2) / ( 1 + n^2 x^2 )^2

lim (n -> +oo) f_n'(x) = (1 - n^2 x^2) / ( 1 + 2n^2 x^2 + n^4 x^4 )

Multiplying both numerator and denominator by 1/n^4 yields

lim (n -> +oo) f_n'(x) = ( 0 - 0) / ( 0 + 0 + x^4 ) = 0 / x^4 = 0.
Since for a large value of x, I can make my n larger than x.

Hence, f(x) = 0 when x = 0 and f(x) = 1, when x is not 0.

Since the function f(x) is not continuous on the interval (-oo, +oo), then
{f_n'(x)} does not converge uniformly on that interval.



Have I done my solution correctly?
 
13
0
It looks correct.
 

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