# Thread: find the value for which the function is continuous

1. ## find 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?

2. 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

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

3. $\displaystyle f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } 0<x not equal to 1 \\c\textrm{ if }x=1\end{cases}$

thats how it is i dont know how to do a not equal sign

4. Originally Posted by dat1611
$\displaystyle f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } x>0 \, ; \, x \ne 1 \\c\textrm{ if }x=1\end{cases}$
fify

5. Originally Posted by dat1611
$\displaystyle f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } 0<x not equal to 1 \\c\textrm{ if }x=1\end{cases}$

thats how it is i dont know how to do a not equal sign
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}$$Once you find that, you can just set c equal to it. From the definitions of continuous functions, f(x) is continuous at point a if \displaystyle \displaystyle \lim\limit_{x \to a}f(x) = f(a)$$

6. ok can someone help with the actual problem

7. Do you know L'Hopital's rule? If you do, just apply to the numerator and denominator and it ' should be pretty easy.

8. Hello, dat1611!

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

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

9. so
$\displaystyle$\displaystyle \lim\limit_{x \to 1}\frac{\ln{x}}{x - 1} = 1