The given limit problem is \(\displaystyle \lim_{x \to 0} \left(\frac{a^x – b^x}{x}\right) \), where \( a, b > 0 \). In this post, we will learn how to solve this limit question.
Solution:
\( \displaystyle \lim_{x \to 0} \left(\frac{a^x – b^x}{x}\right) \)
Rewrite the limit using exponentials:
\( \displaystyle \lim_{x \to 0} \left(\frac{e^{x \ln(a)} – e^{x \ln(b)}}{x}\right) \)
Use the Taylor series expansion for \( e^{x \ln(a)} \) and \( e^{x \ln(b)} \):
\( \displaystyle e^{x \ln(a)} = 1 + x \ln(a) + \frac{x^2 \ln^2(a)}{2} + \cdots \\ \displaystyle e^{x \ln(b)} = 1 + x \ln(b) + \frac{x^2 \ln^2(b)}{2} + \cdots \\ \)Substitute into the limit:
\( \displaystyle \lim_{x \to 0} \frac{\left(1 + x \ln(a) + \frac{x^2 \ln^2(a)}{2} + \cdots \right) – \left(1 + x \ln(b) + \frac{x^2 \ln^2(b)}{2} + \cdots\right)}{x} \\ \)Simplify: (Numerator terms containing \( x^2 \) or more power will be zero, so they can be discarded.)
\( \displaystyle \lim_{x \to 0} \frac{x \ln(a)\, – x \ln(b) + x^2 \left(\cdots \right)}{x} = \ln(a)\, – \ln(b) \\ \\ \)Therefore,
\[ \fbox{$ \lim_{x \to 0} \left(\frac{a^x – b^x}{x}\right) = \ln\left(\frac{a}{b}\right) $} \]Please let me know in the comments if you find any error in this solution.