I'm working on Proofs for Limits. So my question is how the bloody hell do I do this.
This is the equation I am working with:
F(x)= (x^2-5x+8)(x-3)/(x-3)
Use the limit properties to prove algebraically that limf(x)=2 as X approaches 3
I would find the lim x>3-, and the lim x>3+, and show that they are both 2. If they are equal, then
Lim x>3 is 2
Do this by inserting an (x+e) or (x-e) for x, where e is an arbritrary small number that can vanish. In both cases, you will have a e/e in the function, which divides to 1.
I've got a graph. It shows Limf(x) is two, There is some long way thanks though.
To prove algebraically that the limit of f(x) as x approaches 3 is equal to 2, we need to show that:
lim(x→3) f(x) = 2
First, let's simplify the given expression f(x):
F(x) = (x^2 - 5x + 8)(x - 3)/(x - 3)
Since the denominator (x - 3) appears in both the numerator and denominator, we can simplify the expression to:
F(x) = x^2 - 5x + 8
Now, let's find the limit of F(x) as x approaches 3 to show that it equals 2.
lim(x→3) F(x) = lim(x→3) (x^2 - 5x + 8)
To evaluate this limit, we can directly substitute the value x = 3 into the expression:
F(3) = 3^2 - 5(3) + 8
= 9 - 15 + 8
= 17 - 15
= 2
Therefore, we have shown that F(3) = 2.
Since the limit of F(x) as x approaches 3 is equal to the value of F(x) at x = 3, we can conclude that:
lim(x→3) f(x) = 2
This completes the algebraic proof that the limit of f(x) as x approaches 3 is equal to 2.