Final answer:
To prove the continuity of the function f(x) = x²/5x-9 at x₀ = 2, we need to show that for any positive ε, there exists a positive δ such that whenever |x - 2| < δ, we have |f(x) - f(2)| < ε. By calculating f(2) and manipulating the expression |f(x) - f(2)|, we can choose δ = min{1, (2ε)/3} and show that the condition holds for any positive ε. Thus, f(x) is continuous at x₀ = 2.
Step-by-step explanation:
The ε−δ definition of continuity states that a function f is continuous at a point x₀ if for any positive number ε, there exists a positive number δ such that whenever |x - x₀| < δ, we have |f(x) - f(x₀)| < ε.
To prove that the function f(x) = x²/5x-9 is continuous at x₀ = 2, we need to show that for any positive ε, there exists a positive δ such that whenever |x - 2| < δ, we have |f(x) - f(2)| < ε.
Let's calculate f(2):
f(2) = (2²)/(5(2) - 9) = 4/(10 - 9) = 4/1 = 4.
Now, let's assume that |x - 2| < δ, where δ is a positive number. Then we can write:
|f(x) - f(2)| = |(x²)/(5x-9) - 4| = |(x² - 4(5x-9))/(5x-9)|.
By factoring the numerator, we get:
|(x - 2)(x + 2)/(5x-9)|.
Since |x - 2| < δ, we know that |x - 2| < 1, which means that x - 2 < 1 and x + 2 > 3.
Therefore, we can write:
|x - 2| < 1 --> -1 < x - 2 < 1 --> 1 < x + 1 < 3.
Now, let's consider the denominator |5x - 9|:
|5x - 9| = |5(x - 2) + 1|.
Since |x - 2| < 1, we know that -1 < x - 2 < 1, which means that -5 < 5(x - 2) < 5.
Therefore, we can write:
-5 < 5(x - 2) < 5 --> -5 + 1 < 5(x - 2) + 1 < 5 + 1 --> -4 < |5x - 9| < 6.
So, we have:
|f(x) - f(2)| = |(x - 2)(x + 2)/(5x-9)| < 6/(-4) = -3/2.
Since ε is a positive number, we have |f(x) - f(2)| < 3/2 < ε.
Thus, we can choose δ = min{1, (2ε)/3}, and for any positive ε, whenever |x - 2| < δ, we have |f(x) - f(2)| < ε. Therefore, the function f(x) = x²/5x-9 is continuous at x₀ = 2.