## Math stupid question on continuous curve

**category:**general [glöplog]

im so shit at this stuff :(

[url]http://en.wikipedia.org/wiki/Discontinuity_(mathematics)[/]

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How a function with 1/x can "look" continuous for 0 ?

Not to muddle up the good explanation that's already been provided, but I do have a small addition. What you are witnessing (informally called a "hole", as ryg pointed out) occurs any time the numerator and denominator of a rational function are both equal to zero for some value of x (e.g. f(x) = x(x + 1) / x). This is the only circumstance under which this occurs (afaik).

More math threads!

sin(x)/x is not a rational function.

try to represent this function f where f(x) = 1 if x is a rationnal number, and 0 if not.

Lotsa holes ! \o/

Lotsa holes ! \o/

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sin(x)/x is not a rational function.

Good call, do you know of a more accurate term for a function of the form f(x)/g(x) that includes trig functions?

The de L'Hôpital rule: when both the numerator and denominator have the same limit, the limit of the ratio is the ratio of the derivatives.

(sin(4x))'=4cos(x)

x'=1

Therefore sin(4x)/x -> 4cos(x) = 4*1 =4 when x tends to +-0.

(sin(4x))'=4cos(x)

x'=1

Therefore sin(4x)/x -> 4cos(x) = 4*1 =4 when x tends to +-0.

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try to represent this function f where f(x) = 1 if x is a rationnal number, and 0 if not.

Lotsa holes ! \o/

That's plain discontinuities though, not singularities, and the function is well-defined over all reals.

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Good call, do you know of a more accurate term for a function of the form f(x)/g(x) that includes trig functions?

The next larger family with a well-known name would be meromorphic functions, which are quotients of holomorphic (=complex differentiable almost everywhere) functions.

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The de L'Hôpital rule: when both the numerator and denominator have the same limit, the limit of the ratio is the ratio of the derivatives.

(sin(4x))'=4cos(x)

x'=1

Therefore sin(4x)/x -> 4cos(x) = 4*1 =4 when x tends to +-0.

Not exactly.

1. L'Hopital's rule states that when the limit of f(x) = 0 or +-inf and the limit of g(x) = 0 or +-inf, then the limit of f(x)/g(x) is equal to the

*limit of*the ratio of the derivatives.

2. d/dx (sin(4x)) = 4cos(4x)

3. (If we're evaluating the limit as x approaches infinity) The limit of sin / cos is undefined as x approaches infinity, so L'Hopital's rule isn't much help here.

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The next larger family with a well-known name would be meromorphic functions, which are quotients of holomorphic (=complex differentiable almost everywhere) functions.

Cool, thank you. I'll be looking into that.

**Quote:**

The next larger family with a well-known name would be meromorphic functions, which are quotients of holomorphic (=complex differentiable almost everywhere) functions.

Cool, thank you. I'll be looking into that.

orbitaldecay: thanks for pointing the error sin(4x)' is not 4cos(x) but 4cos(4x).

But note that krabob raised the case where x tends to 0, not towards infinity.

I explicated the de l'Hôpital following suggestion by iq.

But note that krabob raised the case where x tends to 0, not towards infinity.

I explicated the de l'Hôpital following suggestion by iq.

You don't need l'Hôpital to get the limit for x->+-inf in this case.

Since |sin(x)|<=1 for all x, lim x->+inf |sin(x)/x| <= lim x->+inf 1/|x| = lim x->inf 1/x = 0. So the limit must be 0.

Since |sin(x)|<=1 for all x, lim x->+inf |sin(x)/x| <= lim x->+inf 1/|x| = lim x->inf 1/x = 0. So the limit must be 0.

The two limits are supposed to be as x->+-inf (doesn't matter which).