3
$\begingroup$

How do I evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L'Hopital or series?

I've tried expanding the variable such as $x = 2y$ or $x = 3y$, but seemed to still get stuck.

$\endgroup$

3 Answers 3

2
$\begingroup$

Once we've proven $\lim_{x\to0}\frac{\sin x}{x}=1$ and $\lim_{x\to0}\frac{x-\sin x}{x^3}=\frac16$ (see e.g. here and here for solutions using neither L'Hôpital's rule nor series), your limit is $\lim_{x\to0}\frac{x-\sin x}{x^3}\frac{x^2}{\sin x}=0$.

$\endgroup$
2
$\begingroup$

First, read this answer. This answer shows that, for $x$ close to $0$, we have the following:

$$\sin x \leq x \leq \tan x$$

Thus, since $\sin x \leq x$ and $\sin x$ has the same sign as $x$ (i.e. either both are positive, both are negative, or both are $0$), we know that $\frac{1}{\sin x} \geq \frac{1}{x}$. Thus, $\frac{1}{\sin x}-\frac{1}{x}\geq 0$.

Also, since $x \leq \tan x$ and $\tan x$ has the same sign as $x$, we know that $\frac{1}{x} \geq \frac{1}{\tan x}$. This implies that $\frac{1}{\sin x}-\frac{1}{x} \leq \frac{1}{\sin x}-\frac{1}{\tan x}$,

Now, we have:

$$0 \leq \frac{1}{\sin x}-\frac{1}{x} \leq \frac{1}{\sin x}-\frac{1}{\tan x}$$

Simplify the trig expressions:

$$0 \leq \frac{x-\sin x}{x\sin x} \leq \frac{1-\cos x}{\sin x}$$

Now, $\lim_{x\to 0} 0=0$ for obvious reasons. Thus, let's consider $\lim_{x\to 0} \frac{1-\cos x}{\sin x}$. Multiply numerator and denominator by $1+\cos x$:

$$\lim_{x\to 0}\frac{1-\cos^2 x}{\sin x(1+\cos x)}=\lim_{x\to 0}\frac{\sin^2 x}{\sin x(1+\cos x)}=\lim_{x\to 0}\frac{\sin x}{1+\cos x}=\frac{\sin 0}{1+\cos 0}=0$$

Thus, $\lim_{x\to 0} 0=\lim_{x\to 0} \frac{1-\cos x}{\sin x}=0$, so by Squeeze Theorem, $\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=0$.

$\endgroup$
0
$\begingroup$

Try the following

$$L=\lim \dfrac{x-x\cos\frac{x}{2}+x\cos\frac{x}{2}-\sin x}{x\sin x}$$ then use the $\sin(x)=2\sin \frac{x}{2}\cos\frac{x}{2}$ now the first two terms limit is zero and the second two terms have a relation with the limit $L$ $$\lim \frac{x \cos\frac{x}{2} -2 \sin\frac{x}{2} \cos\frac{x}{2}}{2x \sin\frac{x}{2} \cos\frac{x}{2}}$$

$$\lim \frac{2 \cos\frac{x}{2}\left( \frac{x}{2}- \sin\frac{x}{2}\right)}{2x \sin\frac{x}{2} \cos\frac{x}{2}}$$

$$\lim \frac{1}{2}\frac{\frac{x}{2}-\sin \frac{x}{2}}{\frac{x}{2}\sin\frac{x}{2}}$$ and the last limit is $1/2L$ so $L=1/2L$ hence $L=0$ by this method you can solve something like $\frac{\sin x -x}{x^3}$ but you need to add and subtract a certain term can figure it out ?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .