#### ccc

Joined Jul 14, 2006
2
It asks me to find the value of c for which the function defined below is continuous. What kind of discontinuity is present if c does not have this value?

ln (x)
------ if 0 < x (not=) 1
x - 1

c, if x = 1

#### Papabravo

Joined Feb 24, 2006
16,162
I'm confused, c does not appear in the function, or am I missing something?

#### Dave

Joined Nov 17, 2003
6,970
Papabravo said:
I'm confused, c does not appear in the function, or am I missing something?
I too am confused. Traditionally the constant c arises due to integration, if so the equation would become:

[ln(x)/(x-1)] + c

But then the subsequent question makes little sense. Can you clarify what you mean?

Dave

#### kinyo

Joined Jun 6, 2005
13
when x=1, the function takes the form 0 divided by 0 ... which is considered undefined ... however, the function approaches the value of 1 within the vicinity of x=1 ... so i believe c must have the value of 1 for the function to be continuous

#### Dave

Joined Nov 17, 2003
6,970
From my interpretation, the function is continuous for x > 0, x /= 1 (where as you correctly state is an undefined state).

But that doesn't explain what c is.

Dave

#### kinyo

Joined Jun 6, 2005
13
c is the value of the function that will make it continuous at x=1 ... the function is said to be discontinuous because it is undefined at x=1, where it results to 0/0

i sound like i'm just repeating the original post ... so let me give another example ... the function (x^2 - 1)/(x-1)

(x^2 - 1)/(x-1) is discontinuous at x=1 as well because it becomes 0/0, so we ask what is the value of the function (call it c) so that it becomes continuous at x=1? as it turns out, the value of c in this case is 2 .. and its easy to figure that out because (x^2 - 1)/(x-1) is simply x + 1, which will have a value of 2 when x=1

and to make things clearer, the problem has nothing to do with the constant of integration which incidentally is also called c in possibly all textbooks in calculus

hth

#### Papabravo

Joined Feb 24, 2006
16,162
To evaluate the limit of an indeterminate form use L'hopital's rule. Take the derivative of the numerator and the derivative of denominator. Now evaluate that quotient.

Edit (Thanks Dave)
Rich (BB code):
d/dx(ln(x)) / d/dx(x-1) = (1/x) /1  = 1/x evaluated at x = 1 is just 1
so c = 1 is the value of the function which makes the original function continuous.

#### Dave

Joined Nov 17, 2003
6,970
Papabravo said:
To evaluate the limit of an indeterminate form use L'hopital's rule. Take the derivative of the numerator and the derivative of denominator. Now evaluate that quotient.

Rich (BB code):
d/dx(ln(x)) / d/dx(x-1) = (1/x) /x  = 1/x^2 evaluated at x = 1 is just 1
so c = 1 is the value of the function which makes the original function continuous.
To correct the above:

d/dx(ln(x)) / d/dx(x-1) = (1/x) /1

However, based this approach it still produces c = 1.

Dave