Show, using limits, that f(x) = x2 – x + 3, is continuous at x = 2.
limit as x->2- = 5
limit as x->2+ = 5
since one-sided limits are the same, and f(2)=5, f is continuous
Do you have to do the delta-epsilon limit proof?
To show that a function f(x) is continuous at a specific point, in this case, x = 2, we need to prove three things:
1. The function is defined at x = 2.
2. The limit of the function exists as x approaches 2.
3. The value of the function at x = 2 is equal to the limit.
Now, let's go step by step to prove these three conditions:
1. The function is defined at x = 2:
Plugging x = 2 into the function f(x) = x^2 - x + 3:
f(2) = 2^2 - 2 + 3 = 4 - 2 + 3 = 5. So the function is indeed defined at x = 2.
2. The limit of the function exists as x approaches 2:
To find the limit of f(x) as x approaches 2, we need to evaluate the following limit:
lim(x->2) (x^2 - x + 3).
We can compute the limit by substituting the value x = 2 directly into the function since it is continuous at that point:
lim(x->2) (x^2 - x + 3) = f(2) = 5.
Thus, the limit of the function as x approaches 2 exists and is equal to 5.
3. The value of the function at x = 2 is equal to the limit:
Since we have already determined that f(2) = 5, and the limit of f(x) as x approaches 2 is also 5, we can conclude that f(x) is continuous at x = 2.
Therefore, using limits, we have shown that f(x) = x^2 - x + 3 is continuous at x = 2.