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If $g(x)=\frac{x}{[x]}$ for $\mathrm{x}>2$, then $\lim _{x \rightarrow 2} \frac{g(x)-g(2)}{x-2}$ is equal to
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$\frac{1}{2}$
Given, $g(x)=\frac{x}{[x]}$
When, $x>2$, then $[x]=2 \Rightarrow g(x)=\frac{x}{2}$
Now, $\quad \lim _{x \rightarrow 2} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{\frac{x}{2}-1}{x-2}$ $=\lim _{x \rightarrow 2} \frac{x-2}{2(x-2)}=\lim _{x \rightarrow 2}\left(\frac{1}{2}\right)=\frac{1}{2}$
When, $x>2$, then $[x]=2 \Rightarrow g(x)=\frac{x}{2}$
Now, $\quad \lim _{x \rightarrow 2} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{\frac{x}{2}-1}{x-2}$ $=\lim _{x \rightarrow 2} \frac{x-2}{2(x-2)}=\lim _{x \rightarrow 2}\left(\frac{1}{2}\right)=\frac{1}{2}$
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