Correct Answer - Option 4 :
\(\frac{m^2}{2}\)
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 \(\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{1 - \cos mx}}{{{x^2}}}} \right]\)
As we can see that, \(\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{1 - \cos mx}}{{{x^2}}}} \right] = \frac{0}{0}\)
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 \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
⇒ \(\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{1 - \cos mx}}{{{x^2}}}} \right] = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{m\sin mx}}{{2x}}} \right] = \frac{0}{0}\)
Again by using L-Hospital's Rule we get,
⇒ \(\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{m\sin mx}}{{2x}}} \right] = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{m^2}\cos mx}}{2}} \right] = \frac{{{m^2}}}{2}\)
⇒ \(\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{1 - \cos mx}}{{{x^2}}}} \right] = \frac{m^2}{2} \)
Hence, option D is the correct answer.
