# Find $\lim_{x \to 0}(e^x-1)(\frac{1}{x} - \left \lfloor{{\frac{1}{x}}}\right \rfloor)$

Find $\lim_{x \to 0}(e^x-1)(\frac{1}{x} - \left \lfloor{{\frac{1}{x}}}\right \rfloor)$

I was thinking about using the Sandwich Theorem and doing something like this: $$(e^x-1) < (e^x-1)(\frac{1}{x} - \left \lfloor{{\frac{1}{x}}}\right \rfloor) < (e^x-1)\cdot \frac{1}{x}$$ (this seems true because $x \to 0$)
and then I can say that the limit of the left side is $0$ and the limit of the right side is $0$ (because I get $=0$ both when $x \to 0^+$ and when $x \to 0^-$)...
So I get the the limit of the original expression equals $0$. But I think this is not correct... Can someone tell me what is a correct way to solve it using the Sandwich Theorem (or something even simpler, without using L'hospital's rule)?

• $0\leq\frac{1}{x} - \left \lfloor{{\frac{1}{x}}}\right \rfloor<1$ Commented Dec 12, 2015 at 16:30
• Limit of right hand side of inequaity is $1$ by L'hospital's rule. Commented Dec 12, 2015 at 16:30
• And the inequality i.e. the left side is wrong. Check for $x=\frac{1}{100000}$. Commented Dec 12, 2015 at 16:32
• @Aniket and Guest - Thank you! Commented Dec 12, 2015 at 16:56

Observe that $0\leq z-\lfloor z\rfloor <1$ so that
$\vert (e^x-1)(\frac{1}{x} - \left \lfloor{{\frac{1}{x}}}\right \rfloor)\vert \leq e^x-1\Rightarrow \lim_{x \to 0}(e^x-1)(\frac{1}{x} - \left \lfloor{{\frac{1}{x}}}\right \rfloor)=0$
• I think that in order for your inequality to be true for all $x$, you need to add an absolute value. Also the limit will be $0$, not $1$. Commented Dec 12, 2015 at 16:38
• Thanks! Can you explain how $z-\lfloor z\rfloor$ can be equal to $1$ ? Commented Dec 12, 2015 at 16:56
• @Natalia Actually the inequality should have been $0 \le z-\lfloor z \rfloor < 1$. Commented Dec 12, 2015 at 16:59
• it is never equal to 1 but that does not matter. < implies $\leq$ but I will change it anyway. Thanks Commented Dec 12, 2015 at 17:00