$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital

Question

$$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$

I tried using $\lim\limits_{x\to0} \frac{\sin x}x=1$.

But it doesn't work :/

Answer 1

$$\frac{x - \sin(x)}{x - \tan(x)} = \frac{x - \sin(x)}{x^3} \cdot \frac{x^3}{x - \tan(x)}$$

Let $x = 3y$ and $x\to 0 \implies y\to 0$ $$\lim_{x\to0} \frac{x - \sin(x)}{x^3} = L $$ $$L = \lim_{y\to0}\frac{3y - \sin(3y)}{(3y)^3} = \lim_{y\to0} \frac 3 {27} \frac{y - \sin(y)}{y^3} + \lim_{y\to0} \frac{4}{27} \frac{\sin^3(y)}{y^3} = \frac{1}{9} L + \frac 4{27} $$

This gives $$\lim_{x\to0}\frac{x - \sin(x)}{x^3} = \frac 1 6 $$

\begin{align*} L &= \lim_{y\to0}\frac{ 3y - \tan(3y)}{27 y^3} \\ &= \lim_{y\to0} \frac{1}{(1 - 3\tan^2(y ))} \cdot \frac{3y(1 - 3\tan^2(y )) - 3 \tan(y) + \tan^3(y)}{27y^3}\\ &= \lim_{y\to0} \frac{1}{(1 - 3\tan^2(y ))} \cdot \left( \frac 3 {27} \frac{y - \tan(y)}{y^3} + \frac 1 {27} \frac{\tan^3(y)}{y^3} - \frac 9 {27} \frac{y \tan^2(y)}{y^3 } \right )\\ &= \frac {3L}{27} + \frac 1 {27} - \frac 1 3 \\ \end{align*}

This gives other limit to be $-1/3$, put it up and get your limit.

Answer 2

Hint

Use the Taylor series: $$\sin x=x-\frac{x^3}{6}+o(x^3)\quad \text{and}\quad \tan x=x+\frac{x^3}{3}+o(x^3)$$

Answer 3

$$ L=\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=\lim_{x\to0}\frac{x-\sin x}{x\cos x-\sin x}\cos x = \lim_{x\to0}\frac{2x-\sin2x}{2x\cos2x-\sin2x}\cos2x\\ = \lim_{x\to0}\frac{x-\cos x\sin x}{x(1-2\sin^2x)-\cos x\sin x}\cos2x=\lim_{x\to0}\frac{x-\cos x\sin x}{x-\cos x\sin x-2x\sin^2x}\cos2x $$ Which, noting that $\lim_{x\to0}\cos2x=1$, we can then write as $$ \lim_{x\to0}\frac{1}{1-\frac{2x\sin^2x}{x-\cos x\sin x}} = \frac{1}{1-2\lim_{x\to0}\frac{x\sin^2x}{x-\cos x\sin x}}=\frac{1}{1-2M} $$ Now, we turn our attention to that new limit... $$ \frac1M=\lim_{x\to0}\frac{x-\cos x\sin x}{x\sin^2x}=\lim_{x\to0}\frac{1-\cos x\frac{\sin x}x}{1-\cos^2x}=1+\lim_{x\to0}\frac{1-\frac{\sin x}{x\cos x}}{1-\cos^2x}\cos^2 x\\ =1+\lim_{x\to0}\frac{x-\tan x}{x\sin^2x} $$ But we also have $$ \frac1M = \lim_{x\to 0} \frac{2x-\sin2x}{2x\sin^2x}=2\lim_{x\to 0} \frac{2x-\sin2x}{2x(1-\cos2x)}=2\lim_{x\to0}\frac{x-\sin x}{x(1-\cos x)}\\ =2\lim_{x\to0}\frac{x-\sin x}{x(1-\cos^2 x)}(1+\cos x)=4\lim_{x\to0}\frac{x-\sin x}{x(1-\cos^2 x)}=4\lim_{x\to0}\frac{x-\sin x}{x\sin^2x}\\ =4\lim_{x\to0}\frac{x-\sin x}{x-\tan x}\cdot\frac{x-\tan x}{x\sin^2x} $$ So, we have $$ \frac1M = 4L\left(\frac1M-1\right) $$ or $1=4L(1-M)$... but $L=\frac{1}{1-2M}$ (or $1=L(1-2M)$).

Therefore, we have that $$ 1-2=4L-4LM-2L+4LM = 2L = -1 $$ Therefore, $L=-\frac12$. No use of $\lim_{x\to0}\frac{\sin x}x$ required.

Answer 4

In the beginning of this answer, it is shown that $$ \begin{align} \frac{\color{#C00000}{\sin(2x)-2\sin(x)}}{\color{#00A000}{\tan(2x)-2\tan(x)}} &=\underbrace{\color{#C00000}{2\sin(x)(\cos(x)-1)}\vphantom{\frac{\tan^2(x)}{\tan^2(x)}}}\underbrace{\frac{\color{#00A000}{1-\tan^2(x)}}{\color{#00A000}{2\tan^3(x)}}}\\ &=\hphantom{\sin}\frac{-2\sin^3(x)}{\cos(x)+1}\hphantom{\sin}\frac{\cos(x)\cos(2x)}{2\sin^3(x)}\\ &=-\frac{\cos(x)\cos(2x)}{\cos(x)+1}\tag{1} \end{align} $$ Therefore, $$ \lim_{x\to0}\,\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}=-\frac12\tag{2} $$ Thus, given an $\epsilon\gt0$, we can find a $\delta\gt0$ so that if $|x|\le\delta$ $$ \left|\,\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}+\frac12\,\right|\le\epsilon\tag{3} $$ Because $\,\displaystyle\lim_{x\to0}\frac{\sin(x)}{x}=\lim_{x\to0}\frac{\tan(x)}{x}=1$, which are shown geometrically in this answer, we have $$ \sin(x)-x=\sum_{k=0}^\infty2^k\sin(x/2^k)-2^{k+1}\sin(x/2^{k+1})\tag{4} $$ and $$ \tan(x)-x=\sum_{k=0}^\infty2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1})\tag{5} $$ By $(3)$ each term of $(4)$ is between $-\frac12-\epsilon$ and $-\frac12+\epsilon$ of the corresponding term of $(5)$.

Therefore, if $|x|\le\delta$ $$ \left|\,\frac{\sin(x)-x}{\tan(x)-x}+\frac12\,\right|\le\epsilon\tag{6} $$ We can restate $(6)$ as $$ \lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)}=-\frac12\tag{7} $$