estimate the maximum error of taylor approximation of $\ln(x)$ find the error bound using the Lagrange Remainder:
$$\ln(x)\approx (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}, |x-1|<\dfrac{1}{64}$$
My attempt:
\begin{align}
R_3(x)&=\dfrac{f^{(4)}(c)}{4!}(x-1)^4\\
f^{4}(c)&=\dfrac{-6}{(x-1)^4}\\
\implies R_3(x)&=\dfrac{-6}{4!}\dfrac{(x-1)^4}{(c-1)^4}\\
\implies |R_3(x)|&\leq \dfrac{6}{4!}\dfrac{(\frac{1}{64})^4}{(\frac{-1}{64})^4}=.25
\end{align}
Is this correct?

Note that,
\begin{equation}
f(x) = log(x) \implies f^{(4)}(x) = \frac{-6}{x^4}
\end{equation}
Therefore,
\begin{equation}
R_3(x) = -\frac{6}{4!c^4}(x-1)^4\quad; \quad \text{where } c \text{ is in between 1 and x}
\end{equation}
Since $c$ depends on $x$, first find an upper bound as;
\begin{equation}
|R_3(x)| \leq
\begin{cases}
\frac{6}{4!}(x-1)^4 \quad &if \quad x>1\\
\frac{6}{4!x^4}(x-1)^4 \quad &if \quad x<1
\end{cases}
\end{equation}
Now we can bound for $|x-1|<1/64$,
\begin{equation}
|R_3(x)| \leq
\begin{cases}
\frac{6}{4!64^4} \quad &if \quad x>1\\
\frac{6}{4!(1-1/64)^464^4} \quad &if \quad x<1
\end{cases}
\end{equation}
Here is a plot done by wolframalpha.com.
You can visaully see that we need two different bounds for $x<1$ and $x>1$. If you need a single bound, $x<1$ case will work.
