shwanky
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I'm a little nervous about a test I have on Thursday and I was wondering if this is adequate for a proof of the following equation.
Show that f(x) is continuous at a=4
f(x) = x^2 + \sqrt{7-x}
For f(x) to be continuous at a = 4
a) f(a) must be defined
b) \lim_{x \to a} f(x) must exist
c) \lim_{x \to a} f(x) = f(a)
Proof
1. f(a) = f(4) = 4^2 + \sqrt{7-4} = 16 + \sqrt{3}
Therefore, f(a) is defined at a=4.
2. \lim_{x \to a} f(x) = \lim_{x \to 4} x^2 + \sqrt{7-x}
\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3}
\lim_{x \to 4^-} x^2 + \sqrt{7-x} = 16 + \sqrt{3}
Since \lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3} = \lim_{x \to 4^-} x^2 + \sqrt{7-x} = \lim_{x \to 4} f(x) exists.
\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}
3. By steps 1 and 2 we know that f(4) = 16 + \sqrt{3} and \lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}.
Since \lim_{x \to 4} f(x) = 16 + \sqrt{3} = f(4), f(x) = x^2 + \sqrt{7-x} is continuous at a = 4.
Homework Statement
Show that f(x) is continuous at a=4
Homework Equations
f(x) = x^2 + \sqrt{7-x}
The Attempt at a Solution
For f(x) to be continuous at a = 4
a) f(a) must be defined
b) \lim_{x \to a} f(x) must exist
c) \lim_{x \to a} f(x) = f(a)
Proof
1. f(a) = f(4) = 4^2 + \sqrt{7-4} = 16 + \sqrt{3}
Therefore, f(a) is defined at a=4.
2. \lim_{x \to a} f(x) = \lim_{x \to 4} x^2 + \sqrt{7-x}
\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3}
\lim_{x \to 4^-} x^2 + \sqrt{7-x} = 16 + \sqrt{3}
Since \lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3} = \lim_{x \to 4^-} x^2 + \sqrt{7-x} = \lim_{x \to 4} f(x) exists.
\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}
3. By steps 1 and 2 we know that f(4) = 16 + \sqrt{3} and \lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}.
Since \lim_{x \to 4} f(x) = 16 + \sqrt{3} = f(4), f(x) = x^2 + \sqrt{7-x} is continuous at a = 4.
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