## ParthKohli 2 years ago Is there a removable discontinuity when a limit exists?

1. abb0t

a removable discontinuity is one at which the limit of the function exists but does not equal the value of the function at that point; this may be because the function does not exist at that point. |dw:1356113778802:dw|

2. ParthKohli

What if there's a hole at a point and the limit *is* the hole? What discontinuity is that called?

3. TuringTest

...but whether or not the unction value exists, for a removeable discontinuity we have that$\lim_{x\to a^1}f(x)=\lim_{x\to a^+}f(x)\neq f(a),\pm\infty$i.e. the limit must exist for a removeable discontinuity.

4. abb0t

Step Continuity?

5. ParthKohli

Oh, I get it.

6. TuringTest

the thing you describe parth is a removable discontinuity

7. ParthKohli

Ah, okay :)

8. TuringTest

the limit exists (is finite) but is not the same as the value of the function at that point. the actual value of the function at that point can be anything... even f(a)=infinity, it would still be removable

9. ParthKohli

Got it, so the discontinuity in a function$f(x) = {{\rm foo} \over x + 2}$is a removable discontinuity right?

10. TuringTest

no, because $\lim_{x\to0^-}f(x)\neq\lim_{x\to0^+}f(x)$

11. TuringTest

I mean $$x\to-2$$

12. TuringTest

also, both limits do not exist; for a removable discontinuity the limit at the discontinuity must exist ( be finite and equal on both sides)

13. TuringTest

here $$x\to2^-\implies f(x)\to\infty,~~~x\to2^+\implies f(x)\to-\infty$$

14. ParthKohli

Ah! Okay, so the limit doesn't exist.

15. TuringTest

so the limits are neither equal on both sides, or even the same sign!

16. TuringTest

correct, for many reasons it doesn't exist

17. TuringTest

it is an essential discontinuity

18. ParthKohli

hmm

19. TuringTest

|dw:1356114605529:dw|

20. ParthKohli

I see :)

21. TuringTest

as $$x\to-2^+$$ we are going this way off to infinity|dw:1356114734440:dw|

22. ParthKohli

Yup, I see it now :) thanks

23. TuringTest

that alone is enough to say that the discontinuity is essential, and that the limit does not exist at x=-2 but still we have the other argument as well. from the left we have the opposite case|dw:1356114780902:dw|

24. ParthKohli

Aww, no one cares about the vertical asymptote $$x = -2$$... give it some love :)

25. TuringTest

forever alone I guess :P

26. ParthKohli

lol

27. ParthKohli

Thanks @TuringTest :)

28. TuringTest

welcome :D