# 1.8.6 AP calculus Exam continuity

• MHB
• karush
In summary, the two expressions are not exactly the same because the second one has a defined value at x=2 while the first one does not. This is due to the fact that the denominator, x-2, cancels out in the first expression, making it undefined at x=2. The second expression, however, remains defined at x=2. This is an important distinction in the definition of continuity, which states that a function is continuous at a point if the limit of the function and the value of the function at that point are equal. In this case, the limit exists, but the value at x=2 does not, making the function not continuous at that point.
karush
Gold Member
MHB
View attachment 9447
well just by observation $x-2$ cancels out leaving $2x+1$ plugging in $2$ gives $2(2)+1 = 5$

or is there some more anointed way to do this.

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Not really a fan of this explanation.

I want to know why

$\dfrac{(2x+1)(x-2)}{x-2}$

is NOT EXACTLY THE SAME as

$(2x+1)$

Always thought filling a hole was more rigorous

definition of continuity at $x=c$ ...

1. $f(c)$ exists

2. $\displaystyle \lim_{x \to c} f(x)$ exists

3. $\displaystyle \lim_{x \to c} f(x) = f(c)$

I shall personally anoint this:
The definition of "continuous" is "f is continuous at x= a if and only if both f(a) and $$\lim_{x\to a} f(x)$$ exist and
$$\lim_{x\to a} f(x)= f(a)$$". Further
$$\lim_{x\to a} f(x)= L$$ is defined by "given any $$\epsilon> 0$$, there exist $$\delta> 0$$ such that if $$0< |x- a|< \delta$$ then $$|f(x)- L|< \epsilon$$.

Note the "$$0< |x- a|$$" which perhaps should be given more emphasis in math classes. The value of $$\lim_{x\to a} f(x)$$ depends entirely upon the values of f(x) for x close to x= a, but the value of f(a), or even whether it exists, is completely irrelevant to the limit!

For this problem, as long as x is not 2, x- 2 is not 0 and the "x- 2" terms in numerator and denominator cancel.

That is, for $$x\ne 2$$, $$f(x)= \frac{(2x+1)(x- 2)}{x- 2}= 2x+1$$.

tkhunny said:
Not really a fan of this explanation.

I want to know why

$\dfrac{(2x+1)(x-2)}{x-2}$

is NOT EXACTLY THE SAME as

$(2x+1)$
They are "NOT EXACTLY THE SAME" because the second has value 5 when x= 2 and the first is not defined there. The graph of y= 2x+ 1 is a straight line while the graph of $$y=\frac{(2x+1)(x- 2)}{x- 2}$$ is that same straight line except for a hole at (2, 5).

## 1. What is continuity in AP Calculus Exam 1.8.6?

Continuity in AP Calculus Exam 1.8.6 refers to a function's ability to be drawn without lifting the pencil from the paper. In other words, the graph of the function is smooth and has no breaks or jumps.

## 2. How is continuity determined in AP Calculus Exam 1.8.6?

In AP Calculus Exam 1.8.6, continuity is determined by three main criteria: the function must exist at the point, the limit of the function at that point must exist, and the limit must be equal to the function value at that point.

## 3. What is the difference between removable and non-removable discontinuity in AP Calculus Exam 1.8.6?

Removable discontinuity in AP Calculus Exam 1.8.6 refers to a point where the function is not continuous but can be made continuous by redefining the function at that point. Non-removable discontinuity, on the other hand, cannot be made continuous by redefining the function at that point.

## 4. How do you determine if a function is continuous at a specific point in AP Calculus Exam 1.8.6?

To determine if a function is continuous at a specific point in AP Calculus Exam 1.8.6, you must check if the three criteria for continuity are met: the function exists at the point, the limit of the function at that point exists, and the limit is equal to the function value at that point.

## 5. Why is continuity important in AP Calculus Exam 1.8.6?

Continuity is important in AP Calculus Exam 1.8.6 because it allows us to determine the behavior of a function at a specific point. It also helps us to understand the relationship between the graph of a function and its algebraic representation.

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