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#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 2,678

- Thread starter
- #3

- Jan 31, 2012

- 2,678

I assume its about k

Always thought filling a hole was more rigorous

Always thought filling a hole was more rigorous

- Mar 1, 2012

- 655

1. $f(c)$ exists

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

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

- Jan 29, 2012

- 1,151

The

[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].

*hole* at (2, 5).

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

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

That is, for [tex]x\ne 2[/tex], [tex]f(x)= \frac{(2x+1)(x- 2)}{x- 2}= 2x+1[/tex].

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 aNot really a fan of this explanation.

I want to know why

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

is NOT EXACTLY THE SAME as

$(2x+1)$