(B) For $x \rightarrow a^{-}$,we have $\frac{x}{a} < 1$,so $\left[\frac{x}{a}\right] = 0$. Thus,$k = \lim _{x \rightarrow a^{-}} \left(\frac{|x|^3}{a}

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(B) For $x \rightarrow a^{-}$,we have $\frac{x}{a} Thus,$k = \lim _{x \rightarrow a^{-}} \left(\frac{|x|^3}{a} - 0^3\right) = \frac{a^3}{a} = a^2$. For $x \rightarrow a^{+}$,we have $\frac{x}{a} > 1$,so $\left[\frac{x}{a}\right] = 1$. Thus,$l = \lim _{x \rightarrow a^{+}} \left(\frac{|x|^3}{a} - 1^3\right) = \frac{a^3}{a} - 1 = a^2 - 1$. Calculating $k - l$: $k - l = a^2 - (a^2 - 1) = a^2 - a^2 + 1 = 1$.

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

Medium

If $a > 0$,$[\cdot]$ denotes the greatest integer function,$\lim _{x \rightarrow a^{-}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=k$,and $\lim _{x \rightarrow a^{+}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=l$,then:

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A
$k=l$
B
$k-l=1$
C
$l-k=1$
D
$l=a^2, k$ does not exist

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Limits and Derivatives

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Concept of limits, Evaluation of algebric limits

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If $a > 0$,$[\cdot]$ denotes the greatest integer function,$\lim _{x \rightarrow a^{-}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=k$,and $\lim _{x \rightarrow a^{+}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=l$,then:

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