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,

$$\lim_{x\to0}\frac{\tan x-x}{x^3}$$

$$\lim_{x\to0}\frac{\sin x-x}{x^3}$$

$$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$

$$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$

$$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$

$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$

Solution 1:

$$L_1=\lim_{x\to0}\frac{\tan x-x}{x^3}\quad L_2=\lim_{x\to0}\frac{\sin x-x}{x^3}\quad L_3=\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}\\L_4=\lim_{x\to0}\frac{e^x-x-1}{x^2}\quad L_5=\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}\quad L_6=\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$

Yes if we know beforehand the limit exists.

For $L_1$: $$L_1=\lim_{x\to0}\frac{\tan x-x}{x^3}\\ L_1=\lim_{x\to0}\frac{\tan 2x-2x}{8x^3}\\ 4L_1=\lim_{x\to0}\frac{\frac12\tan2x-x}{x^3}\\ 3L_1=\lim_{x\to0}\frac{\frac12\tan{2x}-\tan x}{x^3}\\ =\lim_{x\to0}\frac{\tan x}x\frac{\frac1{1-\tan^2x}-1}{x^2}\\ =\lim_{x\to0}\frac{(\tan x)^3}{x^3}=1\\ \large L_1=\frac13$$

For $L_2$: $$L_2=\lim_{x\to0}\frac{\sin x-x}{x^3}\\ L_2=\lim_{x\to0}\frac{\sin 2x-2x}{8x^3}\\ 4L_2=\lim_{x\to0}\frac{\frac12\sin 2x-x}{x^3}\\ 3L_2=\lim_{x\to0}\frac{\frac12\sin 2x-\sin x}{x^3} =\lim_{x\to0}\frac{\cos x-1}{x^2}\frac{\sin x}x\\ L_2=\frac13\lim_{x\to0}\frac{\cos x-1}{x^2}\\ L_2=\frac13\lim_{x\to0}\frac{\cos 2x-1}{4x^2}\\ 4L_2=\frac13\lim_{x\to0}\frac{\cos 2x-1}{x^2}\\ 3L_2=\frac13\lim_{x\to0}\frac{\cos 2x-\cos x}{x^2}\\ 3L_2=\frac13\lim_{x\to0}\frac{-2\sin^2\left(\frac x2\right)(2\cos x+1)}{x^2}\\ 3L_2=\frac13\lim_{x\to0}\frac{-2\sin^2\left(\frac x2\right)(2\cos x+1)}{x^2}\\ \large L_2=-\frac16$$

For $L_3$: $$L_3=\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}\\ L_3=\lim_{x\to0}\frac{\ln(1+2x)-2x}{4x^2}\\ 2L_3=\lim_{x\to0}\frac{\frac12\ln(1+2x)-x}{x^2}\\ L_3=\lim_{x\to0}\frac{\frac12\ln(1+2x)-\ln(1+x)}{x^2}\\ 2L_3=\lim_{x\to0}\frac{\ln(1+2x)-2\ln(1+x)}{x^2}\\ 2L_3=\lim_{x\to0}\frac{\ln\left(1-\frac{x^2}{(1+x)^2}\right)}{x^2}\\ \large L_3=-\frac12 $$

For $L_4$: $$L_4=\lim_{x\to0}\frac{e^x-x-1}{x^2}\\ 4L_4=\lim_{x\to0}\frac{e^{2x}-2x-1}{x^2}\\ 3L_4=\lim_{x\to0}\frac{e^{2x}-e^x-x}{x^2}\\ 12L_4=\lim_{x\to0}\frac{e^{4x}-e^{2x}-2x}{x^2}\\ 6L_4=\lim_{x\to0}\frac{\frac12e^{4x}-\frac12e^{2x}-x}{x^2}\\ 3L_4=\lim_{x\to0}\frac{\frac12e^{4x}-\frac32e^{2x}+e^x}{x^2}\\ 3L_4=\frac12\lim_{x\to0}\frac{e^x(e^x-1)^2(e^x+2)}{x^2}\\ \large L_4=\frac12$$

For $L_5$: $$L_5=\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}\\ 8L_5=\lim_{x\to0}\frac{\sin^{-1}2x-2x}{x^3}\\ 4L_5=\lim_{x\to0}\frac{\frac12\sin^{-1}2x-x}{x^3}\\ 3L_5=\lim_{x\to0}\frac{\frac12\sin^{-1}2x-\sin^{-1}x}{x^3}\\ 6L_5=\lim_{x\to0}\frac{\sin^{-1}2x-2\sin^{-1}x}{x^3}\\ 6L_5=\lim_{x\to0}\frac{\sin^{-1}\left(-4 x^3-2 \sqrt{1-4 x^2} \sqrt{1-x^2} x+2 x\right)}{x^3}\\ 6L_5=\lim_{x\to0}\frac{-4 x^3+2x(1- \sqrt{1-4 x^2} \sqrt{1-x^2})}{x^3}\\ 6L_5=\lim_{x\to0}-4+2\frac{(1- \sqrt{1-5 x^2+4x^4})}{x^2}\\ 6L_5=\lim_{x\to0}-4+2\frac{(1- \sqrt{1-5 x^2+4x^4})}{x^2}$$ Since you would consider binomial theorem as series expansion, if not well and good, if yes, then I'll do: Now let $\sqrt{1-5 x^2+4x^4}=\sum a_kx^k$, squaring both sides, $$1-5x^2+4x^4=a_0^2+2a_0a_1x+(2a_0a_2+a_1^2)x^2+(2a_0a_3+a_1a_2)x^3+(2a_0a_4+2a_1a_3+a_2^2)x^4+...$$ Now taking positive branch: $$a_0=1,a_1=0,a_2=-5/2,a_3=0,a_4=-9/8,...$$ So: $$6L_5=\lim_{x\to0}-4+2\frac{(1- (1-5x^2/2-9x^4/8...))}{x^2}\\\large L_5=\frac16$$

For $L_6$: $$L_6=\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}\\ 4L_6=\lim_{x\to0}\frac{\frac12\tan^{-1}2x-x}{x^3}\\ 3L_6=\lim_{x\to0}\frac{\tan^{-1}2x-2\tan^{-1}x}{2x^3}\\ 6L_6=\lim_{x\to0}\frac{\tan^{-1}\left(-\frac{2 x^3}{3 x^2+1}\right)}{x^3}\\ L_6=-\frac13$$

Solution 2:

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, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\log(1+x)-x}{x^2}=-\frac12}\tag{17} $$

Solution 3:

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.