The \[\mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right]\] (where [x] is greatest integer function) is
Option:
A). A nonzero real number
B). A rational number
C). An integer
D). zero

Hint: In math, a limit can be defined such a way that the function or series tends as the input tends to about some unique value. Limits are very important in the use of calculus and analysis of mathematical modeling, also it is used from different kinds of defined derivatives, continuity, and integrals. Here we have to solve the limit for the greatest integer function.

Complete step-by-step solution:
Step 1:
First we write the given data
 \[\mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right]\]
Now we know that
\[x \geqslant \left[ x \right] \geqslant x - 1\,\,\,\forall \,\,\,x \in R\]
Then we put
\[\Rightarrow x = \dfrac{1}{{{x^3}}}\]
\[\Rightarrow x \geqslant \left[ x \right] \geqslant x - 1\,\,\,\forall \,\,\,x \in R\]
\[\Rightarrow \dfrac{1}{{{x^3}}} \geqslant \left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \dfrac{1}{{{x^3}}} - 1\,\,\,\forall \,\,\,x \in R\]
Then multiply it by \[{x^8}\] so
\[{x^8} \times \dfrac{1}{{{x^3}}} \geqslant {x^8} \times \left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant {x^8} \times \left( {\dfrac{1}{{{x^3}}} - 1} \right)\,\,\,\forall \,\,\,but\]
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^8}}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} {x^{8 - 3}} = \mathop {\lim }\limits_{x \to 0} {x^5}\]
Then put the limit
\[ = \mathop {\lim }\limits_{x \to 0} {x^5} = {0^5} = 0\]
So take the limits below
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} {x^8}\dfrac{1}{{{x^3}}} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8} \times \left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{x^8}}}{{{x^3}}} - {x^8}} \right)\,\,\,\forall \,\,x \in R\]
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^8}}}{{{x^3}}} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^8}}}{{{x^3}}} - \mathop {\lim }\limits_{x \to 0} {x^8}\,\,\,\forall \,\,x \in R\]
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} {x^5} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \mathop {\lim }\limits_{x \to 0} {x^5} - \mathop {\lim }\limits_{x \to 0} {x^8}\,\,\,\forall \,\,x \in R\]
Now put the limits then we got
\[\Rightarrow {0^5} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant {0^5} - {0^8}\,\,\,\forall \,\,x \in R\]
\[\Rightarrow 0 \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant 0 - 0\,\,\,\forall \,\,x \in R\]
\[\Rightarrow 0 \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant 0\,\,\,\forall \,\,x \in R\]
Or we can write it as
\[\mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] = 0\]
So the correct option are
b) A rational number
c) An integer
d) Zero

Note: The given options are correct for the following reasons, option (b) is correct due to the A Rational number due to the ratio of the power are different and put as the integer in the numerator as well as a denominator so the option is correct and then the option (c) is correct due to the in the numerator and the denominator the having the power when we put the integer than integer power is also integer hence the option (c) is correct and the last and final option (d) is correct due to our solution that provided above lead the limiting value of the given question is zero hence the option (d) is correct.