By Continuity definition 1/0 is infinity

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  • Thread starter NotASmurf
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  • #1
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Main Question or Discussion Point

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

Answers and Replies

  • #2
pwsnafu
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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).
 
  • #3
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a is 0, f=1/x
 
  • #4
pwsnafu
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a is 0, f=1/x
Yeah, sorry. I realised and edited my post. You replied just before I finished.
 
  • #5
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oh, 1/x, x element of R, and inf not element of R but cardinality, thanks.
 
  • #6
pwsnafu
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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.
 
  • #7
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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?
 
  • #8
PeroK
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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##
 
  • #9
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but for continous 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.
 
  • #10
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lim 1/x as x->0 is infinity, but the function taking it to infinity is continuous, but for continous 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##.
 

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