Show that _ x 2^- x [x] _ x 2^+ x [a] . - Vidyadip

Question

Show that $\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{[\text{x}]}\ne\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}}{[\text{a}]}.$

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Answer

$\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{[\text{x}]}$ $=\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{1}=\frac{2}{1}=2$ $\bigg[\because\lim\limits_{\text{x}\rightarrow\text{k}^-}\ [\text{x}]=\text{k}-1\bigg]$ Also, $\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}}{[\text{x}]}=\lim\limits_{\text{x}\rightarrow2^-}\frac{\text{x}}{3}=\frac23$ $\bigg[\because\lim\limits_{\text{x}\rightarrow\text{k}^+}\ [\text{x}]=\text{k}+1\bigg]$ $\Rightarrow\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{[\text{x}]}\ne\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}}{[\text{x}]}$

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