Consider a function
that satisfies the limit relation
![Rendered by QuickLaTeX.com \[ \lim_{x \to 0} \left( 1 + x + \frac{f(x)}{x} \right)^{\frac{1}{x}} = e^3, \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e6ee9d74b4d1918c208381eca0843f98_l3.png)
and has a continuous third derivative everywhere. Determine the values
![Rendered by QuickLaTeX.com \[ f(0), \quad f'(0), \quad f''(0), \quad \text{and} \quad \lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-229facb492d9ba91f7815280e4f9ae3c_l3.png)
We do some simplification to the expression first.
![Rendered by QuickLaTeX.com \begin{align*} &&\lim_{x \to 0} \left( 1 + x + \frac{f(x)}{x} \right)^{\frac{1}{x}} &= e^3 \\[9pt] \implies && \lim_{x \to 0} e^{\frac{1}{x} \log \left( 1 + x + \frac{f(x)}{x} \right)} &= e^3 \\[9pt] \implies && \exp \left( \lim_{x \to 0} \left( \frac{1}{x} \log \left( 1 +x + \frac{f(x)}{x} \right) \right) &= e^3 \\[9pt] \implies && \lim_{x \to 0} \left( \frac{1}{x} \log \left( 1 + x + \frac{f(x)}{x} \right) \right) &= 3. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ec0b4bdffe01df4ee9d8f071c57185f4_l3.png)
Now, we apply the hint (that if
then
),
![Rendered by QuickLaTeX.com \begin{align*} &&\frac{1}{x} \log \left( 1 + x + \frac{f(x)}{x} \right) &= 3 + o(1) \\[9pt] \implies && \log \left( 1 + x + \frac{f(x)}{x} \right) &= 3x + o(x) \\[9pt] \implies && 1 + x + \frac{f(x)}{x} &= e^{3x + o(x)} \\[9pt] \implies && x + x^2 + f(x) &= xe^{3x}e^{o(x)} \\[9pt] \implies && f(x) &= - x - x^2 + xe^{3x}e^{o(x)}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-be2fe8c2e09940d9906e4149299d33f2_l3.png)
So as
we have
![Rendered by QuickLaTeX.com \[ f(x) = -x - x^2 + x\left( 1 + 3x + o(x) \right) = 2x^2 + o(x^2). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f49ef55f176d578995bd716caa2bc561_l3.png)
But then since
has three derivatives at 0 we know its Taylor expansion at 0 is unique and is given by (Theorem 7.1, page 274 of Apostol)
![Rendered by QuickLaTeX.com \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2} x^2 + o(x^2). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a42e04744fd476edbd2b7aa9737eb686_l3.png)
Hence, equating the coefficients of like powers of
we have
![Rendered by QuickLaTeX.com \[ f(0) = 0, \quad f'(0) = 0, \quad f''(0) = 4. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-aaf95076ae3d62e09bc8b4310112c5b5_l3.png)
Next, to compute the limit
![Rendered by QuickLaTeX.com \[ \lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-529a408737d61c78f62fd0fb01b45b5c_l3.png)
we write
![Rendered by QuickLaTeX.com \begin{align*} f(x) = 2x^2 + o(x^2) && \implies && \frac{f(x)}{x} &= 2x + o(x) \\[9pt] && \implies && 1 + \frac{f(x)}{x} &= 1 + 2x + o(x) \\[9pt] && \implies && \log \left(1 + \frac{f(x)}{x} \right) &= \log (1+2x+o(x)). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-83f40b6ea58925a9add91505c79c91ae_l3.png)
Then, using the Taylor expansion of
(page 287 of Apostol) we know as
we have
![Rendered by QuickLaTeX.com \[ \log (1+2x+o(x)) = 2x+o(x) + o(x) = 2x + o(x). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a30a6dc05f43295128500402a80c39f6_l3.png)
Therefore we have
![Rendered by QuickLaTeX.com \begin{align*} \log \left( 1 + \frac{f(x)}{x} \right) = 2x + o(x) && \implies && 1 + \frac{f(x)}{x} &= e^{2x+o(x)} \\[9pt] && \implies && \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}} &= e^{2 + o(1)} \\[9pt] && \implies && \lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}} &= e^2. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c071c84211ce81f7dd59d056be28a134_l3.png)