find the limits of floor function:

Question

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 ?

Answer 1

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 $1$.

Answer 2

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 $1$.

Answer 3

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 $1$.