Question

Find ${lim}_{x\to 3^{+}}\frac{x}{\left[x\right]}$. Is it equal to ${lim}_{x\to 3^{-}}\frac{x}{\left[x\right]}$.

Answer 1

${lim}_{x\to 3^{+}}$

Let $x=3+h\Rightarrow h=x-3$

as $x\to 3^{+}\Rightarrow x>3$ slightly

$\Rightarrow x-3>0$

$\Rightarrow h>0\Rightarrow h\to 0^{+}$

$={lim}_{h\to 0^{+}}\frac{3+h}{[3+h]}$

$={lim}_{h\to 0^{+}}\frac{3+h}{3}=\frac{3+0}{3}=1$

Now ${lim}_{h\to 0^{+}}(2-h)=2$

Now ${lim}_{x\to 3^{-}}\frac{x}{\left[x\right]}$

Let $x=3-h\Rightarrow h=3-x$

as $x\to 3^{-}\Rightarrow x<3$ slightly

$\Rightarrow 3-x>0$

$\Rightarrow h>0\Rightarrow h\to 0^{+}$

$={lim}_{h\to 0^{+}}\frac{3-h}{[3-h]}={lim}_{h\to 0^{+}}\frac{3-h}{2}$

$=\frac{3-0}{2}=\frac{3}{2}$

$\therefore {lim}_{x\to 3^{+}}\frac{x}{\left[x\right]}\ne {lim}_{x\to 3^{-}}\frac{x}{\left[x\right]}$