Note that (x^2 - 2ax + a^2) is just (x-a)^2,
and hence f(x) when x =/= a is (x-a)^2/(x-a), which is (x-a)
This limit does exist as x approaches a.
f(a), also exists, because f(a) is f(x) when x = a, which is 5
However, f(x) is not continuous at x = a. This is because, f(a) = 5, as we have already seen. However, the limit of f(x) approaching a is given by (x-a), which tends to 0 at x approaches a.
So, the limit tending to a is not equal to f(a), and the function is not continuous at this point.
Hence, I and II are true, but III is not.
Let f be defined as follows, where a does not = 0,
f(x) = {(x^2 - 2ax + a^2) / (x-a), if x does not = a
5, if x = a
Which of the following are true about f?
I. lim f(x) exists as x approaches a
II. f(a) exists
III. f(x) is continuous at x = a.
A. None
B. I, II, and III
C. I only
D. II only
E. I and II only.
Not too sure about this one and any help is appreciated!
1 answer