Home » Blog » Compute derivatives of a function given a limit equation that it satisfies

Compute derivatives of a function given a limit equation that it satisfies

by RoRi

Consider a function f(x) that satisfies the limit relation

    \[ \lim_{x \to 0} \left( 1 + x + \frac{f(x)}{x} \right)^{\frac{1}{x}} = e^3, \]

and has a continuous third derivative everywhere. Determine the values

    \[ f(0), \quad f'(0), \quad f''(0), \quad \text{and} \quad \lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}}. \]

We do some simplification to the expression first.

    \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*}

Now, we apply the hint (that if \lim_{x \to 0} g(x) = A then g(x) = A + o(1)),

    \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*}

So as x \to 0 we have

    \[ f(x) = -x - x^2 + x\left( 1 + 3x + o(x) \right) = 2x^2 + o(x^2). \]

But then since f 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)

    \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2} x^2 + o(x^2). \]

Hence, equating the coefficients of like powers of x we have

    \[ f(0) = 0, \quad f'(0) = 0, \quad f''(0) = 4. \]

Next, to compute the limit

    \[ \lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}} \]

we write

    \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*}

Then, using the Taylor expansion of \log(1+x) (page 287 of Apostol) we know as x \to 0 we have

    \[ \log (1+2x+o(x)) = 2x+o(x) + o(x) = 2x + o(x). \]

Therefore we have

    \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*}

3 comments

  1. tom says:

    Would I be correct in assuming the motivation for taking logs, for the limit in the second part, is because 1+f(x)/x equaled the taylor series for e^(2x+o(x)) ? Just making sure I’m not missing a ‘trick’.

    • RoRi says:

      Yeah, I think I did that to avoid dealing with the \frac{1}{x} exponent until the end, since that was the trickiest part in the limit. This whole thing could probably be done without taking logs, but I think it gets sort of messy.

      • tom says:

        Very helpful; thanks. Your solution not only avoids a ‘messy’ (albiet probably interesting) alternative but also remains in the context of taylor series and o-notation. I’m glad so many examples of this type are supplied- better tools probably exist for finding limits, but these really are slightly elegant!

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):