\(\displaystyle\lim_{x\rightarrow 0} \dfrac{a^x-b^x}{e^x-1}\) is equal to :

CONCEPT:

• \(\mathop {\lim }\limits_{x\; \to \;0} \left[ {\frac{{{a^x}\; - \;1}}{x}} \right] = \log a,\;a > 0\)
• \(\mathop {\lim }\limits_{x\; \to \;0} \left[ {\frac{{{e^x} - 1}}{x}} \right] = 1\)
• \(\mathop {\lim }\limits_{x\; \to \;a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right)}}{{\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)}},\;provided\;\mathop {\lim }\limits_{x\; \to a} g\left( x \right) \ne 0\)

CALCULATION:

Here, we have to find the limit of \(\displaystyle\lim_{x\rightarrow 0} \dfrac{a^x-b^x}{e^x-1}\)

The expression \(\frac{{{a^x} - {b^x}}}{{{e^x} - 1}}\) can be re-written as:

\(\Rightarrow \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = \left[ {\frac{{\frac{{{a^x} - 1}}{x} - \frac{{{b^x} - 1}}{x}}}{{\frac{{{e^x} - 1}}{x}}}} \right] \)

Now by applying limits on both the sides of the above equation we get

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = \;\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\frac{{{a^x} - 1}}{x} - \frac{{{b^x} - 1}}{x}}}{{\frac{{{e^x} - 1}}{x}}}} \right] \)

As we know that, \(\mathop {\lim }\limits_{x\; \to \;a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right)}}{{\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)}},\;provided\;\mathop {\lim }\limits_{x\; \to a} g\left( x \right) \ne 0\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\frac{{{a^x} - 1}}{x} - \frac{{{b^x} - 1}}{x}}}{{\frac{{{e^x} - 1}}{x}}}} \right] = \frac{{\left[ {\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x}} \right] - \left[ {\mathop {\lim }\limits_{x \to 0} \frac{{{b^x} - 1}}{x}} \right]}}{{\left[ {\mathop {\lim }\limits_{x \to } \frac{{{e^x} - 1}}{x}} \right]}}\)

As we know that, \(\mathop {\lim }\limits_{x\; \to \;0} \left[ {\frac{{{a^x}\; - \;1}}{x}} \right] = \log a,\;a > 0\) and \(\mathop {\lim }\limits_{x\; \to \;0} \left[ {\frac{{{e^x} - 1}}{x}} \right] = 1\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = \log \frac{a}{b}\)

Hence, option A is true.