Discuss the continuity of function This paper discusses the continuity of function f (x) = LIM (1-x ^ 2n / 1 + x ^ 2n) X. if there are discontinuities, the type of discontinuities is determined n→∞ In the given answer, f (x) = - x | x | > 1 0 |x|=1 x |x|

Discuss the continuity of function This paper discusses the continuity of function f (x) = LIM (1-x ^ 2n / 1 + x ^ 2n) X. if there are discontinuities, the type of discontinuities is determined n→∞ In the given answer, f (x) = - x | x | > 1 0 |x|=1 x |x|

This is an exercise of mathematical outline analysis. In fact, there is a routine to solve this kind of problem, that is, to solve the expression of F (x) by finding the limit first. The steps are as follows:
1. Find LIM (1-x ^ 2n / 1 + x ^ 2n) x, (n - > ∞) first:
F (x) = 0 when x = 0 or x = ± 1
X, when 0 ≤ x ＜ 1 or X ＜ - 1
-X, when - 1 ＜ x ≤ 0 or X ＞ 1 (3 cases in total)
2. Then we look for the breakpoint
From the above interval, we can see that there are three "key points": 0, 1, - 1;
(1) Let's look at 0: from the above interval, we can see that limf (0) = limf (x) (x - > 0 +) = limf (x) (x - > 0 -)
So f (x) is continuous at (- 1,1), and 0 is not a discontinuous point;
(2) Let's look at 1: F (1) = 0, limf (x) (x - > 1 -) = x = 1, limf (x) (x - > 1 +) = - x = - 1
F (1) ≠ limf (x) (x - > 1 -)) ≠ limf (x) (x - > 1 +), so x = 1 is the first kind of discontinuity;
(3) Similarly, - 1: F (- 1) = 0, limf (x) (x - > - 1 -) = x = - 1, limf (x) (x - > - 1 +) = - x = 1
F (- 1) ≠ limf (x) (x - > - 1 -)) ≠ limf (x) (x - > - 1 +), so x = - 1 is the first kind of discontinuity;
3. Conclusion: x = 1 and x = - 1 are the first kind of discontinuities; the continuous intervals of F (x) are (- ∞, - 1), (- 1,1), (1, + ∞)

Continuity of piecewise function at piecewise point
Piecewise function f (x) = x ^ A * sin (1 / x ^ b) b > 0, X is not equal to 0,
f(x)=x x=0.
It is proved that f (x) is differentiable at x = 0 when a > 1

It is proved that: [f (0 + Δ x) - f (0)] / Δ x = [(Δ x) ^ A * sin (1 / x ^ b)] / Δ x = (Δ x) ^ (A-1) * sin (1 / x ^ b) = 0 * sin (1 / x ^ b) = 0. Therefore, when a > 1, the function can be derived: (Δ x) ^ (A-1) * sin (1 / x ^ b) = 0 because: (Δ x) ^ (A-1) tends to zero and sin (1 / x ^ B) is bounded

How to judge the continuity of piecewise function

Using left and right limits, if left and right limits exist and are equal and equal to the value of the original function at that point, they are continuous