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Find the limit as x goes to 0 of (ax – 1) / (bx – 1)

Evaluate the limit for b \neq 1.

    \[ \lim_{x \to 0} \frac{a^x - 1}{b^x - 1}. \]


First we write a^x = e^{x \log a} and b^x = e^{x \log b}. Then we use the expansion of e^x (p. 287), to obtain expansions for a^x and b^x,

    \[ a^x = e^{x \log a} = 1 + (x \log a) + o (x), \qquad b^x = e^{x \log b} = 1 + (x \log b) + o(x). \]

Therefore, we have

    \begin{align*}  \lim_{x \to 0} \frac{a^x - 1}{b^x - 1} &= \lim_{x \to 0} \frac{1 + x \log a + o(x) - 1}{1 + x \log b + o(x) - 1} \\[9pt]  &= \lim_{x \to 0} \frac{x \log a + o(x)}{x \log b + o(x)} \\[9pt]  &= \lim_{x \to 0} \frac{\log a + \frac{o(x)}{x}}{\log b + \frac{o(x)}{x}} \\[9pt]  &= \frac{\log a}{\log b}. \end{align*}

5 comments

  1. Mohammad Azad says:

    For those asking, o(x) is little o notation and it means different things in different cases but here it means *some* function f(x) such the quotient f(x)/x approaches 0 as x approaches 0. The definition in Apostol’s book was stated briefly and then Apostol started abusing the notation without explaining everything.

    • Anonymous says:

      For real!! This is my first exposure to the notation and there are so many things I have been confused about/had to sort out on my own.

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