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?
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
$\displaystyle f(x) = \begin{cases}\ln{\left(\frac{x}{x - 1}\right)}\textrm{ if }x \in (0, 1)\cup (1, \infty)\\c\textrm{ if }x=1\end{cases}$?
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}$$
Once you find that, you can just set c equal to it. From the definitions of continuous functions, f(x) is continuous at point a if
$\displaystyle $\displaystyle \lim\limit_{x \to a}f(x) = f(a) $$
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: .$\displaystyle c \:=\:1$