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If $a>0 .[\cdot]$ denotes greatest integer function and $\lim _{x \rightarrow a^{-}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=k, \lim _{x \rightarrow a^{+}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=l$, then
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The correct answer is:
$k-l=1$
For $x \rightarrow a^{-}$
$\frac{x}{a} < 1 \Rightarrow\left[\frac{x}{a}\right]=0$
$\lim _{x \rightarrow a^{-}} \frac{|x|^3}{a}=\frac{a^3}{a}=a^2$
for $x \rightarrow a^{+}$
$\frac{x}{a}>1 \Rightarrow\left[\frac{x}{a}\right]=1$
$\lim _{x \rightarrow a^{+}} \frac{|x|^3}{a}=\frac{a^3}{a}=a^2$
$\begin{aligned} & k=a^2-0=a^2 \\ & l=\left(a^2-1\right) \\ & k-l=a^2-\left(a^2-1\right) \\ & =1\end{aligned}$
Note: In question, for $l$ also limit is $x \rightarrow a^{-}$which should $x \rightarrow a^{+}$
$\frac{x}{a} < 1 \Rightarrow\left[\frac{x}{a}\right]=0$
$\lim _{x \rightarrow a^{-}} \frac{|x|^3}{a}=\frac{a^3}{a}=a^2$
for $x \rightarrow a^{+}$
$\frac{x}{a}>1 \Rightarrow\left[\frac{x}{a}\right]=1$
$\lim _{x \rightarrow a^{+}} \frac{|x|^3}{a}=\frac{a^3}{a}=a^2$
$\begin{aligned} & k=a^2-0=a^2 \\ & l=\left(a^2-1\right) \\ & k-l=a^2-\left(a^2-1\right) \\ & =1\end{aligned}$
Note: In question, for $l$ also limit is $x \rightarrow a^{-}$which should $x \rightarrow a^{+}$
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