f(x)= ln x/(x-1), if 0<x not =1

c , if x=1

and what kind of discontinuity is present if c does not have this value?

can someone help me with this problem?

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- Aug 2nd 2010, 12:39 PMdat1611find the value for which the function is continuous
f(x)= ln x/(x-1), if 0<x not =1

c , if x=1

and what kind of discontinuity is present if c does not have this value?

can someone help me with this problem? - Aug 2nd 2010, 04:40 PMProve It
This is nearly impossible to read...

Is it

$\displaystyle f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if }x \in (0, 1)\cup (1, \infty)\\c\textrm{ if }x=1\end{cases}$

or

$\displaystyle f(x) = \begin{cases}\ln{\left(\frac{x}{x - 1}\right)}\textrm{ if }x \in (0, 1)\cup (1, \infty)\\c\textrm{ if }x=1\end{cases}$? - Aug 3rd 2010, 12:44 PMdat1611
$\displaystyle f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } 0<x not equal to 1 \\c\textrm{ if }x=1\end{cases}$

thats how it is i dont know how to do a not equal sign - Aug 3rd 2010, 12:54 PMskeeter
- Aug 3rd 2010, 01:45 PMVlasev
It is done by \neq

$\displaystyle \neq$

For the question at hand, you need to determine a value c such that the function is continuous. One way could be to just plug in 1 in the above definition with the natural log, but you cannot do that since you'd get something of the form 0/0. Hence you need to find the limit of $\displaystyle \frac{\ln{x}}{x - 1}$ as x approaches 1. Or more succinctly, find

$\displaystyle $\displaystyle \lim\limit_{x \to 1}\frac{\ln{x}}{x - 1}$$

Once you find that, you can just set c equal to it. From the definitions of continuous functions, f(x) is continuous at point a if

$\displaystyle $\displaystyle \lim\limit_{x \to a}f(x) = f(a) $$ - Aug 3rd 2010, 01:46 PMdat1611
ok can someone help with the actual problem

- Aug 3rd 2010, 01:56 PMVlasev
Do you know L'Hopital's rule? If you do, just apply to the numerator and denominator and it ' should be pretty easy.

- Aug 3rd 2010, 02:13 PMSoroban
Hello, dat1611!

Quote:

$\displaystyle f(x)\;=\; \left\{ \begin{array}{cc}\dfrac{\ln x}{x-1}& x > 0,\;x \ne 1 \\ \\[-3mm]

c & x=1 \end{array} \right $

Find the value of $\displaystyle c$ for which the function is continuous.

To be continuous at $\displaystyle x = 1$, we want: .$\displaystyle \displaystyle{\lim_{x\to1}\frac{\ln x}{x-1} \;=\;c }$

The left side goes to $\displaystyle \frac{0}{0}$ so we can apply L'Hopital.

.$\displaystyle \displaystyle{\lim_{x\to1}\frac{\frac{1}{x}}{1} \;=\;c \quad\Rightarrow\quad \lim_{x\to1}\frac{1}{x} \:=\:c$

Therefore: .$\displaystyle c \:=\:1$

- Aug 3rd 2010, 02:21 PMdat1611
so

$\displaystyle $\displaystyle \lim\limit_{x \to 1}\frac{\ln{x}}{x - 1}$

$ = 1

so the answers 1? - Aug 3rd 2010, 02:28 PMVlasev
Yes!Although, it seems like you don't quite understand why. Do you need help with that?