limx→0 ax−bxex−1 is equal to - Infinity Learn by Sri Chaitanya

A
$\log_{e} (\frac{a}{b})$
B
$\log_{e} (\frac{b}{a})$
C
$\log_{e} a b$
D
$\log_{c} (a + b)$

Solution:

We have,

$\begin{array}{l} lim_{x \to 0} \frac{a^{x} - b^{x}}{e^{x} - 1} = lim_{x \to 0} \frac{(a^{x} - 1) - (b^{x} - 1)}{e^{x} - 1} \\ = lim_{x \to 0} \frac{(\frac{a^{x} - 1}{x}) - (\frac{b^{x} - 1}{x})}{(\frac{e^{x} - 1}{x})} = \frac{\log a - \log b}{1} = \log_{e} (\frac{a}{b}) \end{array}$

Related content

CUET Exam Dates 2024 – Application Form, Fees, Eligibility

CBSE Class 12 IP Answer Key 2024,Informatics Practices Paper Solution For SET 1, 2, 3, 4

CUET UG Cut Off 2024, Category, Universities and Colleges Wise Expected Cut Off

Modal Verbs

Helping Verbs

Letter To Your Friend About Your School Trip

Action Verbs

CUET 2024 – List of Colleges and Participating Universities Accepting CUET Exam Score

SRMJEEE Online Test Series – Practice Papers

KIITEE Sample Papers 2024 – Practice Paper PDF Download