mathop {lim }limits_{x to {a^ + }} left( frac | vedclass

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(B) Given the limit: $\mathop {\lim }\limits_{x 	o {a^ + }} \left( \frac{{|x{|^3}}}{a} - {\left[ {\frac{x}{a}} ight]^3} ight)$.Since $a > 0$,as $

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$a^2 - 1$

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(B) Given the limit: $\mathop {\lim }\limits_{x 	o {a^ + }} \left( \frac{{|x{|^3}}}{a} - {\left[ {\frac{x}{a}} ight]^3} ight)$.Since $a > 0$,as $x 	o a^+$,$|x| = x$.Thus,the expression becomes $\mathop {\lim }\limits_{x 	o a^+} \left( \frac{x^3}{a} - \left[ \frac{x}{a} ight]^3 ight)$.As $x 	o a^+$,$\frac{x}{a} 	o 1^+$.Therefore,$\left[ \frac{x}{a} ight] = 1$ for $x$ slightly greater than $a$.Substituting these values into the limit:$= \frac{a^3}{a} - (1)^3 = a^2 - 1$.

$\mathop {\lim }\limits_{x \to {a^ + }} \left( \frac{{|x{|^3}}}{a} - {\left[ {\frac{x}{a}} \right]^3} \right) \,(a > 0)$ is equal to :- (where $[x]$ is the greatest integer function and $|x|$ is the modulus function)

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