$\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{a^x} - {b^x}}}{{{e^x} - 1}}$ is equal to

${\log _e}\left( {\frac{b}{a}} \right)$

${\log _e}\left( {a + b} \right)$

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

\lim\limits_{x \to 0} \,\,\frac{{{a^x} - {b^x}}}{{{e^x} - 1}}

is equal to

View answer and explanation

Correct option: A

{\log _e}\left( {\frac{a}{b}} \right)

Option A (

{\log _e}\left( {\frac{a}{b}} \right)

) is correct because it is the result of applying L'Hopital's rule or by breaking the limit into standard logarithmic limits.

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