Just missed!

Calculus Level 4

$\large \displaystyle\lim_{x\rightarrow 0}{\left\lfloor \left(1-{e}^x\right)\dfrac{\sin x}{\left| x \right| } \right\rfloor } = \, ?$

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

1 -1 Limit does not exist 0

1 solution

Chew-Seong Cheong
Apr 20, 2016

$\begin{aligned} \lim_{x \to 0} \left \lfloor (1-e^x) \frac{\sin x}{|x|} \right \rfloor & = \begin{cases} \lim_{x \to 0^+} \left \lfloor \color{#3D99F6}{(1-e^x)} \color{#D61F06}{\frac{\sin x}{|x|}} \right \rfloor = \left \lfloor (\color{#3D99F6}{0^-}) (\color{#D61F06}{+1}) \right \rfloor = \lfloor 0^- \rfloor \\ \lim_{x \to 0^-} \left \lfloor \color{#3D99F6}{(1-e^x)} \color{#D61F06}{\frac{\sin x}{|x|}} \right \rfloor = \left \lfloor (\color{#3D99F6}{0^+}) (\color{#D61F06}{-1}) \right \rfloor = \lfloor 0^- \rfloor \end{cases} = \boxed{-1} \end{aligned}$

Could you explain the last step you took?

Andrew Tawfeek - 4 years, 9 months ago

Do you mean $\lfloor 0^- \rfloor = -1$ ? Note that $\lfloor 0.5 \rfloor = 0$ but $\lfloor -0.5 \rfloor = -1$ , therefore $\lfloor -0.000.... \rfloor = -1$ .

Chew-Seong Cheong - 4 years, 9 months ago

Just missed!

1 solution

0 pending reports