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[SOLVED] 1.8.360 AP calculus Exam continuity

karush

Well-known member
Jan 31, 2012
2,678
Capture.PNG
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.
 
Last edited:

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
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

Well-known member
Jan 31, 2012
2,678
I assume its about k

Always thought filling a hole was more rigorous
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655
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

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I shall personally anoint this:
The definition of "continuous" is "f is continuous at x= a if and only if both f(a) and [tex]\lim_{x\to a} f(x)[/tex] exist and
[tex]\lim_{x\to a} f(x)= f(a)[/tex]". Further
[tex]\lim_{x\to a} f(x)= L[/tex] is defined by "given any [tex]\epsilon> 0[/tex], there exist [tex]\delta> 0[/tex] such that if [tex]0< |x- a|< \delta[/tex] then [tex]|f(x)- L|< \epsilon[/tex].

Note the "[tex]0< |x- a|[/tex]" which perhaps should be given more emphasis in math classes. The value of [tex]\lim_{x\to a} f(x)[/tex] 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 [tex]x\ne 2[/tex], [tex]f(x)= \frac{(2x+1)(x- 2)}{x- 2}= 2x+1[/tex].

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 [tex]y=\frac{(2x+1)(x- 2)}{x- 2}[/tex] is that same straight line except for a hole at (2, 5).