f(x) = lim(n→∞)？ √( 1+|x|^(3n))
For | x |
f(x) = lim(n→∞)？ √( 1+|x|^(3n)) = 1
For |x| > 1, because the value of | x | (3n) is close to positive infinity, therefore:
f(x) = lim(n→∞)？ √ (1+| x | (3n)) = +∞
When x = 1 or x =-1, |x| = 1, the value of f(x) is:
f( 1) = lim(n→∞)？ √( 1+ 1^(3n)) = 2
f(- 1) = lim(n→∞)？ √( 1+(- 1)^(3n)) = 0
Therefore, the limit of f(x) at x = 1 exists and is equal to 2, at x =-1 exists and is equal to 0, and at | x | 1 the limits are 1 and +∞ respectively.
Next, we can draw an image of f(x) to see if there are any discontinuities. First, we can mark x =-1 and x = 1 on the number axis, and then draw the image of f(x) on the corresponding interval according to the above limit results. We can find that the value of f(x) jumps at x =-1, so x =-1 is a discontinuous point of f(x).
To sum up, f(x) has discontinuous points at x =-1, and other points are continuous.