Is it possible to compute the Taylor-series of $\ln(x)$ for $x = 0$. I get $f\prime(x) = \frac{1}{x}$ and by plugging $0$ , it is undefined form.
Taylor-series of ln(x)
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sequences-and-series
taylor-expansion
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0About $x=0$ the function $\ln x$ is not differentiable. – 2017-01-24
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2It is also not defined. – 2017-01-24
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0So the question : calculate Limit using Taylor-series $\frac{\ln(x)}{x}$ is wrong? where $\lim{x \to 0}$ – 2017-01-24
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1That limit doesn't exist - $|ln(x)|$ tends to infinty, as does $1/x$, so the limit is not defined. – 2017-01-24
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0The limit does exist -- it is $+\infty$. (The limit has to be as $x\to 0^+$, since $\ln$ is not defined for negative numbers.) – 2017-01-24
1 Answers
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This is not possible since
$$x\mapsto \ln(x)$$
is not defined at $x=0$.
Though you can develop $\ln(x)$ in terms of $x-1$, and then divide by $x$ to get the development of $\ln(x)/x$ that you said you wanted in the comments.
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0In other words, it is $ln(1+x)$ which is analytic in the vicinity of $0$. – 2017-01-24
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0@JeanMarie Yes, exactly. – 2017-01-24