1.8.6 AP calculus Exam continuity

[karush]

View attachment 9447

Spoiler

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.

 

[tkhunny]

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

$(2x+1)$

 

[karush]

I assume its about k

Always thought filling a hole was more rigorous

 

[skeeter]

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

1. $f(c)$ exists

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

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

 

[HallsofIvy]

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.

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).

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