Question

Why is Raoul correct?

Joe and Raoul are having a heated discussion about limits and continuity. Joe thinks that if the right and left limits of a function are equal at x=c, and f(c) exists, then f is continuous at x=c. Raoul thinks Joe is making a mistake. Explain why Raoul is correct.

Answer 1

Read below.

Explanation:

First, let's write this to equations.

`"right and left limits of a function are equal at x=c"`

We are saying that `lim_(x->c^-)f(x)=lim_(x->c^+)f(x)`
This tells us that `lim_(x->c)f(x)` exists.

The question also tells us that `f(c)` exists.

Now, is this enough to tell us that the function is continues when `x=c`?

Now, remember that when we are searching for a limit of a given function, (we have `lim_(x->c)f(x)`), we are asking ourselves, `"What y value are we getting closer and closer to as x approaches a certain value"`, not `"What y value are we getting closer and closer to as x equals a certain value"`

If we were to have a point discontinuity, we would have a valid limit at `x=c` and make `f(c)` exist.

Look below for more explanation.

We see that Joe's statement doesn't guarantee continuity.

Answer 2

We start with two definitions::

Definition 1 : Continuity

A function `f(x)` is continuous at `x=a` if `lim_(x rarr a) f(x)` exists, and::

`lim_(x rarr a) f(x) = f(a)`

Definition 2 : Existence of a Limit

The limit, `lim_(x rarr a) f(x)`, exists, and is equal to `L` if:

`lim_(x rarr a^-) f(x) = L \ \` and `\ \ lim_(x rarr a^+) f(x) = L`

From which we can draw two conclusions:

Corollary 1:

If a function is continuous at `x=a` then :

`lim_(x rarr a^-) f(x) = lim_(x rarr a^+) f(x) = f(a)`

Proof: Follows directly from the definition of existence of a limit.

Corollary 2:

If the limit of a function exists at `x=a` then the function is not necessarily continuous at `x=a`

Proof: Existence of a limit satisfies the first part of the continuity definition, but tells us nothing about the second part of the definition regarding the value of `f(a)`

Joe's statement satisfies the conditions of C2, and as such we cannot infer any information about the value of `f(c)` and therefore we can make no comments regarding continuity at `x=c`, thus Raoul is correct.

An example of function fitting the criteria that Joe describes would be a piecewise function with a removable discontinuity at a single point, example:

`f(x) = { (x^2, x != 0), (2, x=0) :}`

Which has a removable discontinuity at `x=0`

And we have:

`lim_(x rarr 0^-) = lim_(x rarr 0^+) = 0`

However:

`f(0) = 2`

Answer 3

`\`

`\mbox{Please see explanation below.}`

Explanation:

`\`

`\mbox{1) The explanation rests on the result that:}`

`\qquad \mbox{a function} \ f(x) \ \mbox{is continuous at} \ x=c`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \iff`

`\qquad \lim_{x\rightarrow c^{-}} f(x), \lim_{x\rightarrow c^{+}} f(x), \ \mbox{and} \ f(c)`

`\qquad \quad \mbox{all exist and are equal}.`

`\`

`\mbox{2) If they all exist, but even a single pair are unequal,} \ \mbox{then} \ f(x) \ \mbox{is not continuous at} \ x=c.`

`\`

`\mbox{3) So what's missing from the heated discussion given,} \ \mbox{to guarantee continuity at} \ x=c, \ \mbox{is the condition} \ \mbox{that} \ f(c) \ \mbox{be equal to both} \ \mbox{one-sided limits above}.`