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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.$$f(x)=\frac{x \sqrt{x}}{(x-6)^{2}} \quad \text { at } x=36$$

$\lim _{x \rightarrow 36} f(x)=\frac{\lim _{x \rightarrow 36}(x \sqrt{x})}{\lim _{x \rightarrow 36}(x-6)^{2}}=\frac{\lim _{x \rightarrow 36} x \cdot \lim _{x \rightarrow 36} \sqrt{x}}{\lim _{x \rightarrow 36}(x-6) \cdot \lim _{x \rightarrow 36}(x-6)}=$$\frac{36 \sqrt{36}}{(36-6)(36-6)}=\frac{36 \cdot 6}{30 \cdot 30}=\frac{216}{900}=\frac{6}{25}=f(36)$

Precalculus

Chapter 13

Limits and Continuity

Section 3

Continuity

Algebra Topics That are Reviewed at the Start of the Semester

Oregon State University

Harvey Mudd College

Utica College

Lectures

07:16

In mathematics, a continuo…

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Use the definition of cont…

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So we have the function f of X, and we want to determine whether F of X is continuous at X equals 36. So, first of all, let's find out what f of 36 is. Make sure it's defined. So f of 36 equals 36 times. Route 36 over 36 minus six squared. We simplify that. Route 36 is just six. So you have 30 to 6 times six, which is, um, 2 16 over 36 months. Six is 30 30 squared is 900. So we have to 16 over 900 which can be simplified, but it doesn't matter. Next, we have to find out that the limit exists. So the limit as X approaches 36 of F of X. And we don't have to worry about left hand and right hand limits because as the um, the function is the same. It's the same function, um, both to the left of 36 to the right of 36. So it's gonna be approaching the same limit value. And when we substitute in 36 for acts are answer isn't indeterminant. It's not 0/0 or infinity over infinity. It's a constant values are. Limit is just that same constant value. So we've determined that the limit exists and that it's equal to the function. Value F of 36 therefore F of X is continuous at X equals 36.

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