I'm not sure, if it is a senceful question but...

 

I always get angry when someone write something like [math] \lim_{x \rightarrow 2 }\left(\frac{x^2-4}{x-2} \right) =4 [/math] because it's not equal ,it approximates, it's "almost equal". I know, it's mathematicly correct, but something's telling me that the equality is not the best symbol for this.

The question is if the "4" on the right represents number or constant function (and my opinion is it should end in this [math] \lim_{x \rightarrow 2 }\left(\frac{x^2-4}{x-2} \right) =\lim_{x \rightarrow 2 } (4) [/math] formula without other editing, if it is a function...) and if that "4" is a real number, why is there equality while that fraction is undefined at 2?

 

many thanks

 

Pq

Edited December 7, 200817 yr by Petanquell
aaaagh, bad number in the second equality...

a) The 4 in your case is (using the definition of limits of real-valued functions I know) a number.

b) With "it approximates it" and "it is almost equal" you probably mean that for values [math]x \approx 2[/math] the function is almost equal to 4. Strictly speaking that's not what the equation sais. The equation claims that the functional (a functional is an operation that assigns a real value to a function, a typical example being integration) [math] \lim_{x\to 2} (F) [/math] is equal to 4 for [math] F=\frac{x^2 - 4}{x-2} [/math].

I'm not sure, if it is a senceful question but...

 

I always get angry when someone write something like [math] \lim_{x \rightarrow 2 }\left(\frac{x^2-4}{x-2} \right) =4 [/math] because it's not equal ,it approximates, it's "almost equal". I know, it's mathematicly correct, but something's telling me that the equality is not the best symbol for this.

Equality is the best symbol for this expression, not approximately equals. This gets at the heart of the definition of the real numbers. Almost all real numbers (i.e., the irrationals) are defined by limits.

 

In the problem at hand, there is no number between [math] \lim_{x \rightarrow 2 }\left(\frac{x^2-4}{x-2} \right)[/math] and 4: They are exactly equal to one another. The expression [math]\frac{x^2-4}{x-2}[/math] is of course undefined at [math]x=2[/math], but the limit is not. So this problem also gets at the heart of the definition of continuity. A function is continuous at a point if

1. the value of the function is defined at that point and
2. the limit of the function as the argument approaches the point in question from any direction exists and is equal to the function value.

The expression [math]\frac{x^2-4}{x-2}[/math] is of course undefined at [math]x=2[/math]...

 

This isn't quite true, though. Since [math]x^2-4 = (x-2)(x+2)[/math] the [math](x-2)[/math] terms cancel and the expression is simply [math]x+2[/math], which of course is pretty easy to find a limit for.

In the problem at hand, there is no number between [math] \lim_{x \rightarrow 2 }\left(\frac{x^2-4}{x-2} \right)[/math] and 4: They are exactly equal to one another.

I think that sums it up quite nicely.

If two real numbers are not equal then there must be another number between them.

I think that sums it up quite nicely.

If two real numbers are not equal then there must be another number between them.

 

The question is, how can be [math] \frac{x^2-4}{x-2} [/math] equal to 4 when the "range of the fraction" is [math] \left(\infty;4 \right)\cup \left(4;\infty \right) [/math] ?

because that is the value it tends towards when it approaches 4

The question is, how can be [math] \frac{x^2-4}{x-2} [/math] equal to 4 when the "range of the fraction" is [math] \left(\infty;4 \right)\cup \left(4;\infty \right) [/math] ?

[math]\left.\frac{x^2-4}{x-2}\right|_{x=2}[/math] and [math]\lim_{x\to 2}\frac{x^2-4}{x-2}[/math] are different expressions. The former expression is not defined while the latter is defined -- and is equal to 4.

 

The value of a function at a point is equal to the limit of a function as the argument approaches the point in question are equal to one another if and only if the function is continuous at the point in question.

The question is, how can be [math] \frac{x^2-4}{x-2} [/math] equal to 4 when the "range of the fraction" is [math] \left(\infty;4 \right)\cup \left(4;\infty \right) [/math] ?

No one (except Bignose) said it would. In part b) in my previous reply I already stated what went wrong, namely that the limit of a function is (or at least "can be considered as") a functional that assigns a value to a function (if the limit exists), not the function itself. Any reason why you (and everyone else) ignored that?

oh, my mistake

 

many thanks guys