- #1

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

[tex]f(x) = x^2 + \sqrt{7-x}[/tex]

For f(x) to be continuous at a = 4

a) f(a) must be defined

b) [tex]\lim_{x \to a} f(x)[/tex] must exist

c) [tex]\lim_{x \to a} f(x) = f(a)[/tex]

1. [tex]f(a) = f(4) = 4^2 + \sqrt{7-4} = 16 + \sqrt{3}[/tex]

Therefore, f(a) is defined at a=4.

2. [tex]\lim_{x \to a} f(x) = \lim_{x \to 4} x^2 + \sqrt{7-x}[/tex]

[tex]\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]

[tex]\lim_{x \to 4^-} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]

Since [tex]\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)[/tex] exists.

[tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]

3. By steps 1 and 2 we know that f(4) = [tex] 16 + \sqrt{3}[/tex] and [tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex].

Since [tex]\lim_{x \to 4} f(x) = 16 + \sqrt{3} = f(4)[/tex], [tex]f(x) = x^2 + \sqrt{7-x}[/tex] is continuous at a = 4.

## Homework Statement

Show that f(x) is continuous at a=4

## Homework Equations

[tex]f(x) = x^2 + \sqrt{7-x}[/tex]

## The Attempt at a Solution

For f(x) to be continuous at a = 4

a) f(a) must be defined

b) [tex]\lim_{x \to a} f(x)[/tex] must exist

c) [tex]\lim_{x \to a} f(x) = f(a)[/tex]

**Proof**1. [tex]f(a) = f(4) = 4^2 + \sqrt{7-4} = 16 + \sqrt{3}[/tex]

Therefore, f(a) is defined at a=4.

2. [tex]\lim_{x \to a} f(x) = \lim_{x \to 4} x^2 + \sqrt{7-x}[/tex]

[tex]\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]

[tex]\lim_{x \to 4^-} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]

Since [tex]\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)[/tex] exists.

[tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]

3. By steps 1 and 2 we know that f(4) = [tex] 16 + \sqrt{3}[/tex] and [tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex].

Since [tex]\lim_{x \to 4} f(x) = 16 + \sqrt{3} = f(4)[/tex], [tex]f(x) = x^2 + \sqrt{7-x}[/tex] is continuous at a = 4.

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