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Use the definition of continuity and the properties of limits to show that the function is continuous at the given number.$$f(x)=\frac{x^{2}-9}{\left(x^{2}-x-6\right)\left(x^{2}+6 x+9\right)} \quad \text { at } x=2$$

$\lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2} \frac{x^{2}-9}{\left(x^{2}-x-6\right)\left(x^{2}+6 x+9\right)}=$$\frac{\lim _{x \rightarrow 2}\left(x^{2}-9\right)}{\lim _{x \rightarrow 2}\left(x^{2}-x-6\right) \cdot \lim _{x \rightarrow 2}\left(x^{2}+6 x+9\right)}=$$\frac{2^{2}-9}{\left(2^{2}-2-6\right)\left(2^{2}+6 \cdot 2+9\right)}=\frac{-5}{(-4)(25)}=\frac{1}{20}=f(2)$

Precalculus

Chapter 13

Limits and Continuity

Section 3

Continuity

Algebra Topics That are Reviewed at the Start of the Semester

Johns Hopkins University

Harvey Mudd College

Idaho State University

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So we have the function f of X and we want to determine whether or not F of X is continuous at X equals two. So to do that will first find what f of two is. Make sure it exists. So if we just sub student for two, we get two squared minus nine mops. Not sure what happened there. Two squared minus nine over two squared, minus two minus six times two squared. Plus six times too plus nine. So we have four minus nine. And the numerator, which is gonna be negative. Five over two squared is four, um, minus two minus six times four plus six times two is 12 plus nine equal to negative five over four months to his negative too. Minus six is negative. Eight times. Ah, 16 plus nine is 25. So that is going to be equal. Teoh 1/40. Um, right, Yeah, 1/40. So we know effort to exist, first of all, so that's good. Our first criteria is filled. Now we need to find the limit as ah f of X approaches. X equals two. So limit comes X approaches two of f of X now because as X approaches to, um f of X will approach f of to weaken, do the same thing. We did hear of substituting, um, in two for X. So if we said student to we're just gonna get the same value 1/40 and we don't have to worry about left hand and right hand ah, limits because it's the same function on both sides. There's no, uh, jump. So, um, we see that 1/40 obviously equals 1/40 so it is f of X is continuous at X equals to. So remember, if you have a limit, if you could just plug in the value that you ah are approaching into f of X or your function, Um and you don't get something indeterminant like 0/0 or infinity over infinity. If you get an actual value than that's just your limit value and because it equals the function value 1/40 we know it's continuous

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