By Continuity definition 1/0 is infinity

[NotASmurf]

lim 1/x as x->0 is infinity, but the function taking it to infinity is continuous, but for continuous functions f(a)= lim f(x) as x->a, so by defininition 1/0 is infinity, what is wrong with this logic?

 

[pwsnafu]

1/x is not defined at x=0. This means that $f(a)$ does not exist, hence you can't appeal to continuity (continuity requires a function to be defined there).

 

[NotASmurf]

a is 0, f=1/x

 

[pwsnafu]

a is 0, f=1/x

Yeah, sorry. I realized and edited my post. You replied just before I finished.

 

[NotASmurf]

oh, 1/x, x element of R, and inf not element of R but cardinality, thanks.

 

[pwsnafu]

Also, I want to point out

lim 1/x as x->0 is infinity

is not true. The left and right limits are not the same.

 

[NotASmurf]

Also, I want to point out

is not true. The left and right limits are not the same.

Why not? 1/x coming from the negative side to 0 should yield same result as from the positive?

 

[PeroK]

Why not? 1/x coming from the negative side to 0 should yield same result as from the positive?

No.The limit from the left is $-\infty$. For example $1/(-.01) = -100$

 

[mfb]

but for continuous functions f(a)= lim f(x) as x->a

There is no way to extend the definition of f to x=0 in a continuous way.

 

[Mark44]

lim 1/x as x->0 is infinity, but the function taking it to infinity is continuous, but for continuous functions f(a)= lim f(x) as x->a, so by defininition 1/0 is infinity, what is wrong with this logic?

Take a look at the graph of y = 1/x. There is the worst possible kind of discontinuity at x = 0, with $\lim_{x \to 0^-} \frac 1 x = -\infty$ and $\lim_{x \to 0^+} \frac 1 x = +\infty$.