Linked Questions

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

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 ...

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.

One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor? $$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$

Evaluate $$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}$$ Without L'Hopital's Rule $$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=\lim_{x\to 0}\frac{x(1-\frac{\sin x}{x})}{x\sin x}=\lim_{x\to 0}\frac{1-\frac{\sin ...

I'm trying to solve this limit $without$ using L'hopital's Rule or Taylor Series. Any help is appreciated! $$\lim\limits_{x\rightarrow 0^+}{\dfrac{e^x-\sin x-1}{x^2}}$$

find the limit without l'Hôpital and Taylor rule : $$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}=?$$ My Try : $$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}\\=\lim\limits_{x \to 0}\frac{...

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{...

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 ...

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.

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 ...