MHB Vertical Lines of x=5 & x=0 in y=ln(x-5)/x

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The discussion centers on the function y=ln(x-5)/x, questioning the presence of vertical asymptotes at x=5 and x=0. It is clarified that there is indeed a vertical line at x=5 due to the logarithm being undefined there. However, the function is also undefined at x=0, leading to the conclusion that a vertical line should be recognized at this point as well. The function cannot approach x=0, reinforcing the existence of the vertical asymptote. Overall, both x=5 and x=0 are critical points for this function.
Petrus
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Hello MHB,
I did find a problem in internet
$$y=\frac{ln(x-5)}{x}$$
they say there is a vertical line at $$x=5$$ but should there not also be a vertical line at $$x=0$$?

Regards,
$$|\pi\rangle$$
 
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Petrus said:
Hello MHB,
I did find a problem in internet
$$y=\frac{ln(x-5)}{x}$$
they say there is a vertical line at $$x=5$$ but should there not also be a vertical line at $$x=0$$?

Regards,
$$|\pi\rangle$$

$$\ln(x-5) $$ is undefined at $x=0$ actually the function cannot get closer to this line.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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