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I have this limit,

$$ \lim_{b\rightarrow0}b\log b +(1-cb-b)\log(1-cb-b)-(1-cb)\log(1-cb) $$

where $c\in\mathbb{R}$ is a constant.

I know this is going to zero, but I want to know how fast it decreases. For instance, when $c=0$ it becomes

$$ \lim_{b\rightarrow0}b\log b+(1-b)\log(1-b) $$ and here the dominat term is $b\log b$ so it decreases as fast as $b\log b$. So, the dominant term in the case of the first limit should be bounded by $b\log b$ but how I get it, and what if $c$ is not constant anymore, if $c=\frac{1}{b}$, the limit still would vanish, but how fast in this case?

thanks

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    $\begingroup$ If $c$ is a constant you shouldn't say it equals $\frac 1b$ because $b$ is varying. $\endgroup$ Commented May 23, 2018 at 13:14
  • $\begingroup$ The expression has the form $b \log b - b+ O(b^2)$ regardless of the (constant) value of $c$. Even if you allow $c$ to depend on $b$, the dependency $c = 1 / b$ gives an undefind expression $\log 0$ in the argument of the limit. $\endgroup$ Commented May 23, 2018 at 13:16
  • $\begingroup$ thanks @RossMillikan, I was just talking of other case. $\endgroup$
    – Luis GC
    Commented May 23, 2018 at 13:33

1 Answer 1

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We have that

$$b\log b +(1-cb-b)\log(1-cb-b)-(1-cb)log(1-cb)=$$

$$=b\log b+(1-cb-b)(-cb-b+o(b))-(1-cb)(-cb+o(b))=$$$$=b\log b-cb-b+cb+o(b)=b\log b-b+o(b)=$$

$$=b\log b + o(b\log b)$$

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  • $\begingroup$ Thanks @gimusi, can you explain me a little more why $\log(1-cb-b)=-cb-b+o(b)$ please? $\endgroup$
    – Luis GC
    Commented May 23, 2018 at 13:31
  • $\begingroup$ @ARo This is just expanding the Maclaurin series of $\log(1 - x)$ to first order. $\endgroup$ Commented May 23, 2018 at 13:32
  • $\begingroup$ @ARo That's first order expansion as $x\to 0\implies \log(1+x)=x+o(x)$ from Taylor's or from standard limits. $\endgroup$
    – user
    Commented May 23, 2018 at 13:33

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