# find the limits of floor function:

find the limits of floor function:

$$\lim_{x\to 0^+} \left\lfloor \dfrac{x^2-2x}{\ln (1-x)}\right\rfloor$$

My Try :

$$\lim_{x\to 0^+} \dfrac{x^2-2x}{\ln (1-x)} =2$$

but floor function is discontinuous at $$x=2$$ . now what ?

• Hint: Because the function $x^2-2x\over \ln(1-x)$ is $<2$ for all relevant $x$, the answer is $1$. Commented Dec 15, 2017 at 11:27
• @user202729 how prove? Commented Dec 15, 2017 at 11:29

Note that $$\log (1-x)< - x$$ for all $x\in(0,1)$ and hence by dividing the inequality with $\log(1-x)<0$ we have $$\frac{x} {\log(1-x)} > -1$$ and multiplying this by $(x-2)<0$ we get $$\frac{x^{2}-2x}{\log(1-x)}< 2-x<2$$ Since the expression $(x^{2}-2x)/\log(1-x)\to 2$ as $x\to 0^{+}$ it follows from the above inequality that as $x\to 0^{+}$ the expression under limit is equal to $1$ and therefore the desired limit is equal to $1$.

• nice derivation as usual!
– user
Commented Dec 15, 2017 at 13:17
• (+1) I didn't see your answer before I wrote mine (which I have deleted)
– robjohn
Commented Dec 15, 2017 at 15:30
• @robjohn: no worry! This kind of thing has happened with me too many times. For such problems there are not many nice ways to solve. Commented Dec 15, 2017 at 16:48

Let $f(x)=x^2-2x-2\ln(1-x).$

Thus, $$f'(x)=\frac{2x(2-x)}{1-x}>0,$$ which says $f(x)>f(0)=0$, which says $$\frac{x^2-2x}{\ln(1-x)}<2$$ for all $0<x<1$ and our limit is equal to $1$.

• This is wrong. $\ln(1-x) < 0$ for $x \in (0,1),$ so the inequality reverses. This can also be seen if you plot the original function, which is always $\le 2$. Commented Dec 15, 2017 at 12:03
• @gammatester I fixed. Thank you! Commented Dec 15, 2017 at 12:05

We substitute $1-x=e^t$ so $t\to0^-$ as $x\to1^+$ then $$\dfrac{x^2-2x}{\ln (1-x)}=\dfrac{e^{2t}-1}{t}<2$$ with $e^t>1+t$ and $t<0$. This shows the limit must be $1$.