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Compute: \(\rm \lim_{x \rightarrow 0}\frac{a^x -1}{2x } = ?\) where a > 0

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Best answer
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.

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