Correct Answer - Option 2 : 1280/3
CONCEPT:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)
II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)
Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
CALCULATION:
Here, we have to find the value of \(\mathop {\lim }\limits_{x \to 2} \left[ {\frac{{{x^{10}} - 1024}}{{{x^3} - 8}}} \right]\)
As we can see that, \(\mathop {\lim }\limits_{x \to 2} \left[ {\frac{{{x^{10}} - 1024}}{{{x^3} - 8}}} \right] = \frac{0}{0}\)
We also know that, according to L-Hospital's Rule if f(x) and g(x) are two functions such that \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\) then \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
⇒ \(\mathop {\lim }\limits_{x \to 2} \left[ {\frac{{{x^{10}} - 1024}}{{{x^3} - 8}}} \right] = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{10\;{x^9}}}{{3\;{x^2}}}} \right] = \frac{{10}}{3} \times {2^7} = \frac{{1280}}{3}\)
Hence, option B is the correct answer.