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$\lim_{x \to 0}\frac{x- \sin{x}}{x^2}$ [duplicate]

Calculate: $$\lim_{x \to 0}\frac{x- \sin{x}}{x^2}$$ I would like to try but i don't find any idea i don't know how to use Hopital rule i tried to return $\cos$ to $\sin$ But it doesn't work a help
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Limit, solution in unusual way

I have a problem with solution of this limit: $$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}$$ Of course, it's a very easy to solve, using (twice) L'Hôpital's rule, but I need to find out, how to do this ...
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Finding the limit of $(1-\cos(x))/x$ as $x\to 0$ with squeeze theorem

How do I find: $$\lim_{x\to0}\frac{1-\cos(x)}{x}$$ Using the squeeze theorem. Particularly, how would I arrive at its bounding functions? If possible, please try not to use derivatives.
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Evaluation of $\lim\limits_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$

One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor? $$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$
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Trying to find $\lim_{x \to 0} \frac{x - \sin x}{(x \sin x)^{(3/2)}}$ using L'Hopital's
I'm trying to use L'Hopital's rule to calculate: $$\lim_{x \to 0^+} \dfrac{x - \sin x}{(x \sin x)^{(3/2)}}$$ Taking a couple of derivatives of the denominator gets quite nasty, so I'd like to find a ...
I am new to this site, so I don't know if this will appear correctly. I need to solve this limit without L'Hospital's Rule: $$\lim_{x\to 0}\frac{x-\sin x}{x^3}.$$ I know that the result is $\frac{... 0answers 249 views $\lim_{x\to0}\frac{\sin x-x}{x^3}$without de l'Hospital's Rule? [duplicate] I know, how to calculate $$\lim_{x\to0}\frac{\cos x-1}{x^2}$$ without differential calculus. Calculating $$\lim_{x\to0}\frac{\sin x-x}{x^3}$$ using de l'Hospital's rule or Taylor expansion is also ... 1answer 188 views How to solve$\lim\limits_{x\to 0} \frac{x - \tan(x)}{x^2}$Without L'Hospital's Rule? How to solve$\lim\limits_{x\to 0} \frac{x - \tan(x)}{x^2}$Without L'Hospital's Rule? you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff. 3answers 110 views Calculate$\lim_{x\to 0} \frac{x-\sin x} {1-\cos x}\$
Calculate the limit without using de l'Hopital: $$\lim_{x\to 0} \frac{x-\sin x} {1-\cos x}$$ I want to use the limit:$$\lim_{x\to 0} \frac{\sin x}{x}=1$$ but I don't know how to do it. I manipulated ...