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I'm trying to find the following limit (without L'Hopital, not there yet):

$$\lim_{x \rightarrow 0} \frac{x -\sin x}{x (1 - \cos x)}$$

Tanking into account that:

$$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$

and

$$\lim_{x \rightarrow 0} \frac{1 - \cos x}{\frac{x^2}{2}} = 1$$

If I divide by x I get:

$$\lim_{x \rightarrow 0} \frac{x -\sin x}{x (1 - \cos x)} = \lim_{x \rightarrow 0} \frac{1 - \frac{\sin x}{x}}{(1 - \cos x)}$$

And I still get a $\frac{0}{0}$ indeterminate limit. I have also tried dividing numerator by $\frac{x^2}{2}$.

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

But I still get a $\frac{0}{0}$ indeterminate limit. How can I solve this limit problem without L'Hopital?

Jordi
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    Use https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion $$\lim_{x\to0}\dfrac{x-\sin x}{x^3}=?$$ – lab bhattacharjee Feb 13 '24 at 16:08

2 Answers2

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Consider the leading behavior of $\sin(x)$ and $\cos(x)$:

$$\sin(x)=x-\frac{x^3}{3!}+O(x^5)$$ $$\cos(x)=1-\frac{x^2}{2!}+O(x^4)$$

So we have;

$$\lim_{x \rightarrow 0} \frac{x-x+\frac{x^3}{3!}-O(x^5)}{x(1-1+\frac{x^2}{2!}-O(x^4))}$$

$$=\lim_{x \rightarrow 0} \frac{\frac{x^3}{3!}-O(x^5)}{\frac{x^3}{2!}-O(x^5)}$$

$$=\lim_{x \rightarrow 0} \frac{\frac{1}{3!}-O(x^2)}{\frac{1}{2!}-O(x^2)}$$

$$= \frac{\frac{1}{3!}}{\frac{1}{2!}}$$

$$=\frac{1}{3}$$

Volk
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we have $$ I=\lim_{x\to 0} \frac{x-\sin x}{x(1-\cos x)}=\lim_{x\to 0} \frac{(x-\sin x)(1+\cos x)}{x(1-\cos^2 x)}$$ So $$ I=(1+\cos0)\lim_{x\to 0} \frac{x-\sin x}{x\sin^2x}=2\lim_{x\to 0} \frac{x^2}{\sin^2x}\frac{x-\sin x}{x^3}=2\lim_{x\to 0}\frac{x-\sin x}{x^3}$$ where its famous problem which proved here $$ \lim_{x\to0} \frac{x-\sin x}{x^3}=\frac{1}{6}$$ So $$ \lim_{x\to 0} \frac{x-\sin x}{x(1-\cos x)}=\frac{1}{3}$$

Faoler
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