Are all limits solvable without L'Hôpital Rule or Series Expansion Is it always possible to find the limit of a function without using L'Hôpital Rule or Series Expansion?
For example, 
$$\lim_{x\to0}\frac{\tan x-x}{x^3}$$
$$\lim_{x\to0}\frac{\sin x-x}{x^3}$$
$$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$
$$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$
$$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$
$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$
 A: Using only trigonometric identities, in this answer, it is shown that
$$
\lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)}=-\frac12\tag{1}
$$
Therefore, if we subtract from $1$, we get
$$
\lim_{x\to0}\frac{\tan(x)-\sin(x)}{\tan(x)-x}=\frac32\tag{2}
$$
Using the limits proven geometrically in this answer, we can derive
$$
\begin{align}
\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}
&=\lim_{x\to0}\frac{\tan(x)(1-\cos(x))}{x^3}\\
&=\lim_{x\to0}\frac{\tan(x)}x\frac{\sin^2(x)}{x^2}\frac1{1+\cos(x)}\\
&=\frac12\tag{3}
\end{align}
$$
we can divide $(3)$ by $(2)$ to get
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\tan(x)-x}{x^3}=\frac13}\tag{4}
$$
and we can multiply $(1)$ by $(4)$ to get
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\sin(x)-x}{x^3}=-\frac16}\tag{5}
$$
Note that $(4)$ implies
$$
\begin{align}
\lim_{x\to0}\frac{\tan(x)-x}{\tan^3(x)}
&=\lim_{x\to0}\frac{\tan(x)-x}{x^3}\lim_{x\to0}\frac{x^3}{\tan^3(x)}\\
&=\frac13\cdot1\tag{6}
\end{align}
$$
Therefore, substituting $x\mapsto\tan^{-1}(x)$,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\tan^{-1}(x)-x}{x^3}=-\frac13}\tag{7}
$$
Similarly, $(5)$ implies
$$
\begin{align}
\lim_{x\to0}\frac{\sin(x)-x}{\sin^3(x)}
&=\lim_{x\to0}\frac{\sin(x)-x}{x^3}\lim_{x\to0}\frac{x^3}{\sin^3(x)}\\
&=-\frac16\cdot1\tag{8}
\end{align}
$$
Therefore, substituting $x\mapsto\sin^{-1}(x)$,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\sin^{-1}(x)-x}{x^3}=\frac16}\tag{9}
$$

Using the Binomial Theorem, we have
$$
\left(1+\frac xn\right)^n-1-x
=\frac{n-1}{2n}x^2+\sum_{k=3}^n\binom{n}{k}\frac{x^k}{n^k}\tag{10}
$$
and for $|x|\le1$,
$$
\begin{align}
\left|\sum_{k=3}^n\binom{n}{k}\frac{x^k}{n^k}\right|
&=|x|^3\left|\sum_{k=3}^n\binom{n}{k}\frac{x^{k-3}}{n^k}\right|\\
&\le |x|^3\sum_{k=3}^\infty\frac1{k!}\\[6pt]
&=|x|^3\left(e-\tfrac52\right)\tag{11}
\end{align}
$$
Combining $(10)$ and $(11)$ and taking the limit as $n\to\infty$ yields
$$
\frac{e^x-1-x}{x^2}=\frac12+O(|x|)\tag{12}
$$
and therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{e^x-1-x}{x^2}=\frac12}\tag{13}
$$
A simple corollary of $(13)$ is
$$
\lim_{x\to0}\frac{e^x-1}x=1\tag{14}
$$
Therefore, it follows that
$$
\begin{align}
\lim_{x\to0}\frac{e^x-1-x}{(e^x-1)^2}
&=\lim_{x\to0}\frac{e^x-1-x}{x^2}\lim_{x\to0}\frac{x^2}{(e^x-1)^2}\\
&=\frac12\tag{15}
\end{align}
$$
If we substitute $x\mapsto\log(1+x)$ in $(15)$, we get
$$
\lim_{x\to0}\frac{x-\log(1+x)}{x^2}=\frac12\tag{16}
$$
Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\log(1+x)-x}{x^2}=-\frac12}\tag{17}
$$
A: In general, $ \lim_{x \to 0} \frac{f(x) - \sum_{k = 1}^{n - 1} \frac{f^{(k)}(0)\cdot x^k}{k!}}{x^n} = \frac{f^{(n)}(0)}{n!} $. This can be proven using the Mean Value Theorem $n$ times and induction.
