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Find the limit as x goes to 0 of (ax – asin x) / x3

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0} \frac{a^x - a^{\sin x}}{x^3}. \]

First, we want to get expansions for a^x and a^{\sin x} as x \to 0. For a^x we write a^x = e^{x \log a} and use the expansion (page 287 of Apostol) of e^x. This gives us

    \[ a^x = e^{x \log a} = 1 + (x \log a) + \frac{(\log a)^2}{2} x^2 + \frac{(\log a)^3}{6} x^3 + o(x^3). \]

Next, for a^{\sin x}, again we write a^{\sin x} = e^{\sin x \log a} and then use the expansion for e^x we have

    \[ a^{\sin x} = e^{\sin x \log a} = 1 + (\log a \sin x) + \frac{(\log a)^2}{2} (\sin x)^2 + \frac{(\log a)^3}{6} (\sin x)^3 + o((\sin x)^3). \]

Now, we need use the expansion for \sin x (again, page 287 of Apostol)

    \[ \sin x = x - \frac{x^3}{6} + o(x^4) \]

and substitute this into our expansion of a^{\sin x},

    \begin{align*} a^{\sin x} &= 1 + (\log a \sin x) + \frac{(\log a)^2}{2} (\sin x)^2 + \frac{(\log a)^3}{6} (\sin x)^3 + o((\sin x)^3) \\[9pt] &= 1 + (\log a) \left( x - \frac{x^3}{6} + o(x^4) \right) + \frac{(\log a)^2}{2} \left( x - \frac{x^3}{6} + o(x^4) \right)^2 \\ & \qquad + \frac{(\log a)^3}{6} \left( x - \frac{x^3}{6} + o(x^4) \right)^3 + o\left( \left( x - \frac{x^3}{6} + o(x^4) \right)^3 \right) \\[9pt] &= 1 + (\log a) x + \frac{(\log a)^2}{2} x^2 + \left( -\frac{\log a}{6} + \frac{(\log a)^3}{6} \right) x^3 + o(x^3). \end{align*}

(Again, this is the really nice part of little o-notation. We had lots of terms in powers of x greater than 3, but they all get absorbed into o(x^3), so we don’t actually have to multiply out and figure out what they all were. We only need to figure out the terms for the powers of x up to 3. Of course, the 3 could be any number depending on the situation; we chose 3 in this case because we know that’s what we will want in the limit we are trying to evaluate.)

So, now we have expansions for a^x and a^{\sin x} (in which most of the terms cancel when we subtract) and we can evaluate the limit.

    \begin{align*} \lim_{x \to 0} \frac{a^x - a^{\sin x}}{x^3} &= \lim_{x \to 0} \frac{\frac{\log a}{6} x^3 + o(x^3)}{x^3} \\[9pt] &= \lim_{x \to 0} \left(\frac{\log a}{6} + \frac{o(x^3)}{x^3}\right)\\[9pt] &= \frac{\log a}{6}. \end{align*}

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