Evaluate the limit.
![Rendered by QuickLaTeX.com \[ \lim_{x \to 0} \frac{a^x - a^{\sin x}}{x^3}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ff9c286844f18b381309719a8e0f8c68_l3.png)
First, we want to get expansions for
and
as
. For
we write
and use the expansion (page 287 of Apostol) of
. This gives us
![Rendered by QuickLaTeX.com \[ a^x = e^{x \log a} = 1 + (x \log a) + \frac{(\log a)^2}{2} x^2 + \frac{(\log a)^3}{6} x^3 + o(x^3). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e793d98433a3144e64b861e6af793a5a_l3.png)
Next, for
, again we write
and then use the expansion for
we have
![Rendered by QuickLaTeX.com \[ a^{\sin x} = e^{\sin x \log a} = 1 + (\log a \sin x) + \frac{(\log a)^2}{2} (\sin x)^2 + \frac{(\log a)^3}{6} (\sin x)^3 + o((\sin x)^3). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3fe1b9b4b5d8ac946531082212601be9_l3.png)
Now, we need use the expansion for
(again, page 287 of Apostol)
![Rendered by QuickLaTeX.com \[ \sin x = x - \frac{x^3}{6} + o(x^4) \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9531d050eb45776243722bf339614ebb_l3.png)
and substitute this into our expansion of
,
![Rendered by QuickLaTeX.com \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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-19898b8eb582339f5bed6dadf2864efa_l3.png)
(Again, this is the really nice part of little
-notation. We had lots of terms in powers of
greater than 3, but they all get absorbed into
, 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
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
and
(in which most of the terms cancel when we subtract) and we can evaluate the limit.
![Rendered by QuickLaTeX.com \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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f6a1f75851b0dca4943f4995bf1f2802_l3.png)