(C) Given,$g(x) = \frac{x}{[x]}$.For $x > 2$ and $x$ close to $2$,$[x] = 2$,so $g(x) = \frac{x}{2}$.Also,$g(2) = \frac{2}{[2]} = \frac{2}{2} = 1$.Now,

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(C) Given,$g(x) = \frac{x}{[x]}$.For $x > 2$ and $x$ close to $2$,$[x] = 2$,so $g(x) = \frac{x}{2}$.Also,$g(2) = \frac{2}{[2]} = \frac{2}{2} = 1$.Now,we evaluate the right-hand limit:$\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^+} \frac{1}{2} = \frac{1}{2}$.

Class 11 Mathematics Limits Concept of limits, Evaluation of algebric limits

Easy

If $g(x) = \frac{x}{[x]}$ for $x > 2$,then $\lim_{x \rightarrow 2^+} \frac{g(x) - g(2)}{x - 2}$ is equal to

AP EAMCET 2015 All AP EAMCET

A
$-1$
B
$0$
C
$\frac{1}{2}$
D
$2$

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Concept of limits, Evaluation of algebric limits

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If $g(x) = \frac{x}{[x]}$ for $x > 2$,then $\lim_{x \rightarrow 2^+} \frac{g(x) - g(2)}{x - 2}$ is equal to

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