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Find the value of \(\mathop {\lim }\limits_{x \to 2} \left[ {\frac{{{x^{10}} - 1024}}{{{x^3} - 8}}} \right]\) ?
1. 640/3
2. 1280/3
3. 320/3
4. None of these

1 Answer

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Best answer
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.

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