How can I study the limit of the following function? $$\begin{equation*} \lim_{x \rightarrow 0} \frac{[x]}{\sin x} \end{equation*}$$
Any hint will be appreciated!
Thanks!
How can I study the limit of the following function? $$\begin{equation*} \lim_{x \rightarrow 0} \frac{[x]}{\sin x} \end{equation*}$$
Any hint will be appreciated!
Thanks!
Bumbble Comm
On
For $0<x<1 $
$$ \frac{[x]}{\sin x} =0 $$ Hence
\begin{equation*} \lim_{x \rightarrow 0^+} \frac{[x]}{\sin x}=0 \end{equation*} Whereas for $-1<x<0$
$$ \frac{[x]}{\sin x} = \frac{-1}{\sin x} $$
Thus \begin{equation*} \lim_{x \rightarrow 0^-} \frac{[x]}{\sin x}=\infty \end{equation*} does not exist.
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You need to evaluate both right and left hand limits here, as $[x]$ (floor function) is not continuous at $x = 0$:
If $x \in [0,1)$, then $[x] = 0$. So the right hand limit is simply $0$.
If $x \in (-1, 0)$, then $[x] = -1$. The left hand limit is:
$$\lim_{x\to 0^-} \frac{-1}{ \sin(x)} =\infty$$
From here we see the limit does not exist.