Thread:

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

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

or

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

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

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

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

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

thats how it is i dont know how to do a not equal sign
It is done by \neq

[LaTeX ERROR: Convert failed]

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 [LaTeX ERROR: Convert failed] as x approaches 1. Or more succinctly, find

[LaTeX ERROR: Convert failed]

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

[LaTeX ERROR: Convert failed]

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!

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

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

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

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

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

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

9. so
$$\displaystyle \lim\limit_{x \to 1}\frac{\ln{x}}{x - 1}$<br />$ = 1
so the answers 1?

10. Yes!Although, it seems like you don't quite understand why. Do you need help with that?
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9208202958106995, "perplexity_flag": "middle"}