Može li netko pokušati riješiti ovaj limes sa ili bez L'H, svejedno mi je...
[dispmath]\lim_{x\to 0}\left(\frac{1}{x^2}-\mathrm{ctg}^{2}x\right)[/dispmath]

Bez l'Hopital-a nemam ideju kako bi išlo, moguće da čak i ne postoji način.
Evo s l'Hopital-om:
[dispmath]\lim_{x\to 0}\left(\frac{1}{x^2}-\mathrm{ctg}^2 x\right)=\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{\cos^2 x}{\sin^2 x}\right)=\lim_{x\to 0}\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}=\lim_{x\to 0}\frac{\left(\sin^2 x-x^2\cos^2 x\right)'}{\left(x^2\sin^2 x\right)'}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{2\sin x\cos x-\left[2x\cos^2 x+x^2\cdot 2\cos x\left(-\sin x\right)\right]}{2x\sin^2 x+x^2\cdot 2\sin x\cos x}=\lim_{x\to 0}\frac{2\sin x\cos x-2x\cos^2 x+2x^2\sin x\cos x}{2x\sin^2 x+2x^2\sin x\cos x}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{\sin x\cos x-x\cos^2 x+x^2\sin x\cos x}{x\sin^2 x+x^2\sin x\cos x}=\lim_{x\to 0}\frac{\left(\sin x\cos x-x\cos^2 x+x^2\sin x\cos x\right)'}{\left(x\sin^2 x+x^2\sin x\cos x\right)'}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{\cos^2 x-\sin^2 x-\left[\cos^2 x+x\cdot 2\cos x\cdot\left(-\sin x\right)\right]+2x\sin x\cos x+x^2\left(\sin x\cos x\right)'}{\sin^2 x+x\cdot 2\sin x\cos x+2x\sin x\cos x+x^2\left(\sin x\cos x\right)'}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{\cos^2 x-\sin^2 x-\cos^2 x+2x\sin x\cos x+2x\sin x\cos x+x^2\cos^2 x-x^2\sin^2 x}{\sin^2 x+2x\sin x\cos x+2x\sin x\cos x+x^2\cos^2 x-x^2\sin^2 x}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{-\sin^2 x+4x\sin x\cos x+x^2\cos^2 x-x^2\sin^2 x}{\sin^2 x+4x\sin x\cos x+x^2\cos^2 x-x^2\sin^2 x}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{\left(-\sin^2 x+4x\sin x\cos x+x^2\cos^2 x-x^2\sin^2 x\right)'}{\left(\sin^2 x+4x\sin x\cos x+x^2\cos^2 x-x^2\sin^2 x\right)'}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{-2\sin x\cos x+4\sin x\cos x+4x\left(\sin x\cos x\right)'+2x\cos^2 x+x^2\cdot 2\cos x\cdot\left(-\sin x\right)-2x\sin^2 x-x^2\cdot 2\sin x\cos x}{2\sin x\cos x+4\sin x\cos x+4x\left(\sin x\cos x\right)'+2x\cos^2 x+x^2\cdot 2\cos x\cdot\left(-\sin x\right)-2x\sin^2 x-x^2\cdot 2\sin x\cos x}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{2\sin x\cos x+4x\left(\sin x\cos x\right)'+2x\cos^2 x-2x^2\sin x\cos x-2x\sin^2 x-2x^2\sin x\cos x}{6\sin x\cos x+4x\left(\sin x\cos x\right)'+2x\cos^2 x-2x^2\sin x\cos x-2x\sin^2 x-2x^2\sin x\cos x}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{\sin x\cos x+2x\left(\sin x\cos x\right)'+x\cos^2 x-2x^2\sin x\cos x-x\sin^2 x}{3\sin x\cos x+2x\left(\sin x\cos x\right)'+x\cos^2 x-2x^2\sin x\cos x-x\sin^2 x}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{\sin x\cos x+2x\cos^2 x-2x\sin^2 x+x\cos^2 x-2x^2\sin x\cos x-x\sin^2 x}{3\sin x\cos x+2x\cos^2 x-2x\sin^2 x+x\cos^2 x-2x^2\sin x\cos x-x\sin^2 x}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{\sin x\cos x+3x\cos^2 x-3x\sin^2 x-2x^2\sin x\cos x}{3\sin x\cos x+3x\cos^2 x-3x\sin^2 x-2x^2\sin x\cos x}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{\left(\sin x\cos x+3x\cos^2 x-3x\sin^2 x-2x^2\sin x\cos x\right)'}{\left(3\sin x\cos x+3x\cos^2 x-3x\sin^2 x-2x^2\sin x\cos x\right)'}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{\cos^2 x-\sin^2 x+3\cos^2 x-3x\cdot 2\cos x\sin x-3\sin^2 x-3x\cdot 2\sin x\cos x-2\cdot 2x\sin x\cos x-2x^2\left(\sin x\cos x\right)'}{3\cos^2 x-3\sin^2 x+3\cos^2 x-3x\cdot 2\cos x\sin x-3\sin^2 x-3x\cdot 2\sin x\cos x-2\cdot 2x\sin x\cos x-2x^2\left(\sin x\cos x\right)'}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{4\cos^2 x-4\sin^2 x-6x\sin x\cos x-6x\sin x\cos x-4x\sin x\cos x-2x^2\left(\sin x\cos x\right)'}{6\cos^2 x-6\sin^2 x-6x\sin x\cos x-6x\sin x\cos x-4x\sin x\cos x-2x^2\left(\sin x\cos x\right)'}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{4\cos^2 x-4\sin^2 x-16x\sin x\cos x-2x^2\left(\sin x\cos x\right)'}{6\cos^2 x-6\sin^2 x-16x\sin x\cos x-2x^2\left(\sin x\cos x\right)'}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{2\cos^2 x-2\sin^2 x-8x\sin x\cos x-x^2\left(\sin x\cos x\right)'}{3\cos^2 x-3\sin^2 x-8x\sin x\cos x-x^2\left(\sin x\cos x\right)'}=[/dispmath]
[dispmath]=\lim_{x\to 0}\frac{2\cos^2 x-2\sin^2 x-8x\sin x\cos x-x^2\cos^2 x+x^2\sin^2 x}{3\cos^2 x-3\sin^2 x-8x\sin x\cos x-x^2\cos^2 x+x^2\sin^2 x}=[/dispmath]
[dispmath]=\frac{2}{3}[/dispmath]

Jako ti zahvaljujem na trudu jer za ovo je trebalo vremena!

Drugo, ovaj zadatak mi je bio na ispitu, i čak sam bio na dobrom putu, ali sam odustao misleći da nema šanse da bude tako dug... :/ Ne znam koji je cilj ovakvih zadatakam pogotovo kad imaš još 4 teška i sve to u samo 2 sata... Čini mi se da nikad neću proći ovu matematiku

Iako meni ne treba rešenje, za toliki trud i ja moram zahvaliti Danielu.

Inače, kad bih ja ovo dobio na ispitu, pre bih krenuo da uradim razvoj oko nule. Dobije se:
[dispmath]\frac{2}{3}-\frac{x^2}{15}-\frac{2 x^4}{189}+\cdots[/dispmath]
I onda ti rešenje naravno bude [inlmath]\frac{2}{3}[/inlmath].

Pa, da... Ja po sistemu „što da bude jednostavno, kad može komplikovano“...