eseper je napisao:prvo sam se rješio sinusa, pa [inlmath]\ln[/inlmath]-a

Negde u ovom postupku ti je greška, jer izraz koji si posle toga dobio ne teži istoj vrednosti kao i onaj prethodni.

(1)[dispmath]\lim_{x\to 0}\;\;\sin 4x\frac{e^{-2x}-1}{x^2\left(\sqrt{x+1}-1\right)}\ln(1+x)[/dispmath]
(2)[dispmath]=4\lim_{x\to 0}{\cancelto{1}{\frac{\sin 4x}{4x}}}\frac{\ln(1+x)\left(e^{-2x}-1\right)}{x\left(\sqrt{x+1}-1\right)}=[/dispmath]
(3)[dispmath]=4\lim_{x\to 0}\frac{\ln(1+x)^{\frac{1}{x}x}\left(e^{-2x}-1\right)}{x\left(\sqrt{x+1}-1\right)}=[/dispmath]
(4)[dispmath]=4\lim_{x\to 0}\frac{{\cancel{x}}\left(e^{-2x}-1\right)}{{\cancel{x}}\left(\sqrt{x+1}-1\right)}=[/dispmath]
(5)[dispmath]=4\lim_{x\to 0}\frac{\left(e^{-2x}-1\right)}{\left(\sqrt{x+1}-1\right)}=[/dispmath]
(6)[dispmath]=4\lim_{x\to 0}\frac{\left(e^{-2x}-1\right)}{\left(\sqrt{x+1}-1\right)}\frac{\left(\sqrt{x+1}+1\right)}{\left(\sqrt{x+1}+1\right)}=[/dispmath]
(7)[dispmath]=4\lim_{x\to 0}\frac{{\cancelto{2}{\left(\sqrt{x+1}+1\right)}}\left(e^{-2x}-1\right)}{x}=[/dispmath]
(8)[dispmath]=8\lim_{x\to 0}\frac{e^{-2x}-1}{x}=[/dispmath]
(9)[dispmath]=8\lim_{x\to 0}\frac{{\cancelto{-1}{-e^{-2x}}}\left(e^x-1\right){\cancelto{2}{\left(e^x+1\right)}}}{\ln\left(1+e^x-1\right)}=[/dispmath]
(10)[dispmath]=-16\frac{1}{\ln e}=-16[/dispmath]

eseper je napisao:

Bez korištenja L'Hopitalovog pravila rješite limes:

[dispmath]\lim_{x\to 0}\;\;\sin 4x\frac{e^{-2x}-1}{x^2\sqrt{x+1}-1}\ln(1+x)[/dispmath]

Ovde si imenilac napisao bez zagrade...

eseper je napisao:(1)[dispmath]\lim_{x\to 0}\;\;\sin 4x\frac{e^{-2x}-1}{x^2{\color{red}(}\sqrt{x+1}-1{\color{red})}}\ln(1+x)[/dispmath]
(2)[dispmath]\cdots[/dispmath]

... a ovde si radio sa zagradom u imeniocu...

Eto bar ostaje postupak za one koji ne budu znali

Kako rješiti limes
[dispmath]\lim_{x\to\pi}\left(\frac{2\pi x-x^2}{\pi^2}\right)^{\mathrm{ctg}^2x}[/dispmath]

[dispmath]2\pi x-x^2=\pi^2-\left(\pi^2-2\pi x+x^2\right)=\pi^2-\left(\pi-x\right)^2[/dispmath]
Da li ti ovo daje ideju?

I ne baš

OK...[dispmath]\lim_{x\to\pi}\left(\frac{2\pi x-x^2}{{\pi}^2}\right)^{\mathrm{ctg}\:^2x}=\cdots=\lim_{x\to\pi}\left[\frac{\pi^2-\left(\pi-x\right)^2}{{\pi}^2}\right]^{\mathrm{ctg}\:^2x}=\lim_{x\to\pi}\left[1-\left(\frac{\pi-x}{\pi}\right)^2\right]^{\mathrm{ctg}\:^2x}=[/dispmath][dispmath]=\left\{\lim_{x\to\pi}\left[1-\left(\frac{\pi-x}{\pi}\right)^2\right]^{-\left(\frac{\pi}{\pi-x}\right)^2}\right\}^{-\left(\frac{\pi-x}{\pi}\right)^2\mathrm{ctg}\:^2x}=e^{-\lim\limits_{x\to\pi}\left(\frac{\pi-x}{\pi}\right)^2\mathrm{ctg}\:^2x}[/dispmath]
[dispmath]-\lim\limits_{x\to\pi}\left(\frac{\pi-x}{\pi}\right)^2\mathrm{ctg}\:^2x=-\lim\limits_{x\to\pi}\left(\frac{\pi-x}{\pi}\right)^2\frac{\cancelto{1}{\cos^2 x}}{\sin^2 x}=[/dispmath][dispmath]=-\lim\limits_{x\to\pi}\left(\frac{\pi-x}{\pi}\right)^2\frac{1}{\sin^2\left(\pi-x\right)}=-\frac{1}{\pi^2}\lim\limits_{x\to\pi}\left[\cancelto{1}{\frac{\pi-x}{\sin\left(\pi-x\right)}}\right]^2=-\frac{1}{\pi^2}[/dispmath]
[dispmath]\Rightarrow\quad e^{-\lim\limits_{x\to\pi}\left(\frac{\pi-x}{\pi}\right)^2\mathrm{ctg}\:^2x}=e^{-\frac{1}{\pi^2}}[/dispmath]

[dispmath]\lim_{x\to\pm\infty}\frac{|x|\sqrt{1-\frac{3}{x}}}{x}-1[/dispmath][dispmath]\mathrm{sign}(x)=0,\;x\to +\infty[/dispmath][dispmath]\mathrm{sign}(x)=-2,\;x\to -\infty[/dispmath]
Kako se dobiju [inlmath]0[/inlmath] i [inlmath]-2[/inlmath]?

Notacija je bzvz. Ne može nikako biti [inlmath]\mathrm{sign}\left(x\right)=-2[/inlmath], jer funkcija [inlmath]\mathrm{sign}\left(x\right)[/inlmath] po definiciji može imati samo vrednosti [inlmath]-1[/inlmath], [inlmath]0[/inlmath] ili [inlmath]1[/inlmath].

Tu se sigurno misli na to da je [inlmath]\lim\limits_{x\to +\infty}\left(\frac{\left|x\right|\sqrt{1-\frac{3}{x}}}{x}-1\right)=0[/inlmath] i [inlmath]\lim\limits_{x\to -\infty}\left(\frac{\left|x\right|\sqrt{1-\frac{3}{x}}}{x}-1\right)=-2[/inlmath]

A evo kako se došlo dotle: zbog onog [inlmath]\left|x\right|[/inlmath] moramo odvojeno da ispitujemo limes kad [inlmath]x\to +\infty[/inlmath], a odvojeno kad [inlmath]x\to -\infty[/inlmath]. Kada [inlmath]x\to +\infty[/inlmath], onda je, naravno, [inlmath]x[/inlmath] pozitivno, pa je [inlmath]\left|x\right|=x[/inlmath], a kada [inlmath]x\to -\infty[/inlmath], onda je [inlmath]x[/inlmath] negativno, pa je [inlmath]\left|x\right|=-x[/inlmath]:[dispmath]\lim_{x\to +\infty}\left(\frac{\left|x\right|\sqrt{1-\frac{3}{x}}}{x}-1\right)=\lim_{x\to +\infty}\left(\frac{\cancel x\sqrt{1-\frac{3}{x}}}{\cancel x}-1\right)=\lim_{x\to +\infty}\left(\sqrt{1-\cancelto{0}{\frac{3}{x}}}-1\right)=\sqrt 1-1=0[/dispmath]
[dispmath]\lim_{x\to -\infty}\left(\frac{\left|x\right|\sqrt{1-\frac{3}{x}}}{x}-1\right)=\lim_{x\to -\infty}\left(\frac{-x\sqrt{1-\frac{3}{x}}}{x}-1\right)=[/dispmath][dispmath]=\lim_{x\to -\infty}\left(-\frac{\cancel x\sqrt{1-\frac{3}{x}}}{\cancel x}-1\right)=\lim_{x\to -\infty}\left(-\sqrt{1-\cancelto{0}{\frac{3}{x}}}-1\right)=-\sqrt 1-1=-2[/dispmath]