1. ## Limits/continuity

Hi,

First post, was not sure where to start with this one. I'm looking for an example of a function that is defined for all real numbers and for which the limit as x approaches 0 exists, but the limit as x approaches 0 is not the same as when the function f(0).

IIRC this would mean it is discontinuous (removable I think?) so would any function that when direct substitution is used returns 0/0 (and requires factoring, common denominator, conjugate, etc to find its limit) work for this? or would that still not be defined for all real numbers I guess, I'm kind of lost on where to start.

2. ## Re: Limits/continuity

What about \displaystyle \displaystyle \begin{align*} f(x) = \begin{cases} \frac{x^2 + 3x + 2}{x + 1} \textrm{ if } x \neq -1 \\ 5 \textrm{ if } x = -1 \end{cases} \end{align*}

3. ## Re: Limits/continuity

Thanks for the response, this is where I get confused though - the first one would be 2/0 and dividing by 0 makes it undefined, so would that disqualify the "function that is defined for all numbers" part? Also x would be approaching 0, not -1, unless i'm looking at something differently here?

4. ## Re: Limits/continuity

Originally Posted by Holly
Thanks for the response, this is where I get confused though - the first one would be 2/0 and dividing by 0 makes it undefined, so would that disqualify the "function that is defined for all numbers" part? Also x would be approaching 0, not -1, unless i'm looking at something differently here?
It's a hybrid function, we have realised that the first part of the function is undefined at the point x = -1 so we have defined the function to have a value at that point.

And of course, you need to evaluate the limit as x goes to -1, because that's the point of discontinuity.

5. ## Re: Limits/continuity

Is -1 coming from anything specific though or is it just a random number? I know it makes the first 0, but using 0 itself does too, so is there any other reason why -1 is used?

Sorry, just having trouble wrapping my head around these concepts in general.

6. ## Re: Limits/continuity

Originally Posted by Holly
Is -1 coming from anything specific though or is it just a random number? I know it makes the first 0, but using 0 itself does too, so is there any other reason why -1 is used?

Sorry, just having trouble wrapping my head around these concepts in general.
No, x = -1 does not make the first function 0, it makes the DENOMINATOR of the first function 0, which gives the point of discontinuity.

x = 0 does NOT make the denominator 0, so is NOT a point of discontinuity.

7. ## Re: Limits/continuity

The simplest such example is just f(x)= 1 if $\displaystyle x\ne 0$, f(0)= 0.