Let f_n(x) = x/(1+nx^2) for n=natural numbers and x of real numbers. Find pointwise limit of (f_n) on R...?

Question

Is the limit uniform? (Hint: Find the max and min values of f_n). For which values of x is (lim f_n)'(x) = lim f '_n(x)?

Answer 1

The pointwise limit of f_n on R exists if for every real number x, f_n(x) converges as a sequence.

A sequence a_n converges to a value a iff for every epsilon, there is an N such that n > N implies that a_n is within epsilon of a.

Now consider f_n(x). For large enough n, for any non-zero x, nx^2 can be as large as you want, certainly as much bigger than 1 as you want. Therefore, for large enough n, f_n(x) is approximately:

x/(nx^2) = 1/nx

So the only remaining question is what happens at x = 0

Since this is not the answer the question leads one to expect, either:

1. I am off-base
2. The problem is not correctly stated
3. The question is designed to be misleading.

Answer 2

permit f(x) = x/(x-3) discover factors the place f(x) = -4: x = -4(x-3) x = -4x + 12 5x = 12 x = 12/5, so: f(12/5) = -4 we would desire to be attentive to the place interior the x-selection this function produces larger/decrease values. For this we take the by-manufactured from f(x) with the citation rule: f'(x) = (a million.(x-3) - x.a million) / (x-3)² = -3 / (x-3)² this spinoff is often destructive, different than for x=3 the place that's undefined. this implies that f(x) will cut back for increasing x as long as x would not go the cost 3. so this implies that f(x) < -4 for x interior the era (12/5, 3) for x > 3 itis sparkling that f(x) is often useful. end: x < 12/5 or x > 3 nb: the 1st (edit: andsecond) poster made an errors via multiplying with (x-3) which will substitute the inequality (<) while x-3 is destructive. edit: I see 2d poster additionally made an errors. please attempt those recommendations: as an occasion: x = -10 => the priority holds x/(x-3) > -4 x = 13/5 => the priority would not carry x/(x-3) < -4 so x = -10 shoud be interior the form and x = 13/5 not.