Correct Answer - Option 4 :
\(\rm \frac{ log \ a}{2 }\)
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 as:
\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
Calculation:
\(\rm \lim_{x \rightarrow 0}\frac{a^x -1}{2x } = \frac{0}{0}\) here a > 0
So, by using L'Hospitals rule i.e. Differentiating numerator and denominator w.r.t. x.
⇒ \(\rm \lim_{x \rightarrow 0}\frac{a^x -1}{2x } = \rm \lim_{x \rightarrow 0}\frac{a^x log a}{2 }\)
⇒ \(\rm \lim_{x \rightarrow 0}\frac{a^x log a}{2 } = \rm \frac{ log \ a}{2 }\)
Hence, option 4 is the correct answer.