Determining Continuity of a Function Without a Given Point

[ARYT]

OK. Starting with a basic question, can we determine whether a function is continuous in general?
So far, our tutorial questions were all about continuity/ discontinuity at a given point. I mean, we should firstly prove that the right-hand and the left-hand limits are equal (while x tends to c) and then the obtained value should be equaled to the value of function at the given point which is f(c).
For this question we have no “c” in fact. It IS asking for that “c”, to some extent.
The question: For what values of x is the function f(x) continuous?

Cheers

 

[Focus]

OK. Starting with a basic question, can we determine whether a function is continuous in general?
So far, our tutorial questions were all about continuity/ discontinuity at a given point. I mean, we should firstly prove that the right-hand and the left-hand limits are equal (while x tends to c) and then the obtained value should be equaled to the value of function at the given point which is f(c).

Continuous just means it is continuous at every point. The easiest way of proving it is just letting the point be arbitrary.

For this question we have no “c” in fact. It IS asking for that “c”, to some extent.
The question: For what value of x is the function f(x) continuous?

Cheers

Are you trying to figure out the points this function is continuous? If so then as long as you don't divide by zero, you can use the algebra of limits (i.e. if f and g are cts then so are f+g, f*g and f/g given g is non-zero).

 

[ARYT]

Thanks for the response. I was also going to solve the question using an arbitrary value, but it says "For what VALUES of x", so most probably, there should be an interval of continuity for this function.

Let's show some effort. This is my answer:

The denominator shouldn't be zero. So, x cannot be +5 and -5.

The final value for the square root should not be negative. Therefore: x<-3 and x>+3

We can clearly deduct that we have infinite discontinuity in +5 and -5 and jump discontinuity in -3<x<+3.

Thus, the values of x in which f(x) is continuous: (-∞,-5); (-5,-3]; [+3, +5); (+5, +∞)

 

[HallsofIvy]

Your analysis is correct but you should not say it has a "jump discontinuity in -3<x<+3."
The function simply is not defined between -3 and 3.