Find $\lim_{{x \to 0}} \frac{x - \sin x}{x - \tan x}$ without using L'Hôpital's rule

Question

I'm trying to find this limit:

$$\lim_{{x \to 0}} \frac{x - \sin x}{x - \tan x}$$

I want to use only algebraic manipulations and trigonometric identities, without L'Hôpital's rule or Taylor series.

I tried to use all the identities I know but I couldn't get rid of the indeterminate form.

Answer 1

Using Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor, one has $$ \begin{eqnarray} &&\lim_{x\to0}\frac{x-\tan x}{x^3}\\ &=&\lim_{x\to0}\frac1{\cos x}\frac{(x\cos x-x)+(x-\sin x)}{x^3}\\ &=&\lim_{x\to0}\frac1{\cos x}\bigg(\frac{\cos x-1}{x^2}+\frac{x-\sin x}{x^3}\bigg) \\ &=&\lim_{x\to0}\frac1{\cos x}\bigg(\frac{x-\sin x}{x^3}-\frac{2\sin^2(\frac x2)}{x^2}\bigg) \\ &=&\frac16-\frac12=-\frac13 \end{eqnarray} $$ and hence

$$ \lim_{{x \to 0}} \frac{x - \sin x}{x - \tan x}=\lim_{{x \to 0}} \frac{x - \sin x}{x^3}\cdot\frac{x^3}{x - \tan x}=\frac16(-3)=-\frac12.$$$

Answer 2

According to your question, we have to find $$\lim_{x\to0} \frac{x-\sin(x)}{x-\tan(x)}$$. Now the series expansion of $\sin(x)$ is $x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-...$ and the series expansion of $\tan(x)$ is $x+\frac{x^{3}}{3}+\frac{2x^{5}}{15}+...$ Now just put the series expansion of $\sin(x)$ and $\tan(x)$ to find the value of the above limit. So, you will get $$\lim_{x\to0} \frac{x-(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-...)}{x-(x+\frac{x^{3}}{3}+\frac{2x^{5}}{15}+...)}$$. Now take $\frac{x^{3}}{3!}$ common from the numerator and $\frac{-x^{3}}{3}$ common from the denominator. So, the final answer of your limits will be $\frac{-1}{2}$.