# how fast decrease a limit

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

• If $c$ is a constant you shouldn't say it equals $\frac 1b$ because $b$ is varying. Commented May 23, 2018 at 13:14
• 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. Commented May 23, 2018 at 13:16
• thanks @RossMillikan, I was just talking of other case. Commented May 23, 2018 at 13:33

We have that

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

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

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

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