Correct Answer - Option 3 : 3/5
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 \(\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^{3x} - 1}}}{{5x}}} \right]\)
As we can see that, \(\mathop {\lim }\limits_{x \to 0 } \left[ {\frac{{{e^{3x} - 1}}}{{5x}}} \right] = \frac{0}{0}\)
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 \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
⇒ \(\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^{3x}} - 1}}{{5x}}} \right] = \;\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{3{e^{3x}}}}{5}} \right] = \frac{3}{5}\)
Hence, option C is the correct answer.