[dispmath]\begin{array}{ll}
\displaystyle\lim_{x\to0}\frac{\mathrm{tg}\:x-\sin x}{x^3}\quad & \displaystyle\left(\frac{1}{2}\right)\\
\\
\displaystyle\lim_{x\to0}\frac{1-\cos^3x}{x^2}\quad & \displaystyle\left(\frac{3}{2}\right)
\end{array}[/dispmath]
može li pomoć sa ova dva limesa? u zagradi su navedena rješenja

Evo prvog, uskoro će i drugi...
[dispmath]\lim_{x\to0}\frac{\text{tg }x-\sin x}{x^3}=\lim_{x\to0}\frac{\text{tg }x}{x}\cdot\frac{1-\cos x}{x^2}=\lim_{x\to0}\frac{1}{\cos x}\cdot\frac{\sin x}{x}\cdot\frac{1-\cos x}{x^2}=[/dispmath][dispmath]=\lim_{x\to0}\frac{1}{\cos x}\cdot\lim_{x\to0}\frac{\sin x}{x}\cdot\lim_{x\to0}\frac{1-\cos x}{x^2}=1\cdot1\cdot\lim_{x\to0}\frac{1-\cos x}{x^2}=[/dispmath][dispmath]=\lim_{x\to0}\frac{2\sin^2\frac{x}{2}}{4\left(\frac{x}{2}\right)^2}=\frac{2}{4}\lim_{x\to0}\frac{\sin^2\frac{x}{2}}{\left(\frac{x}{2}\right)^2}=\frac{1}{2}\lim_{x\to0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2=\frac{1}{2}\cdot1^2=\frac{1}{2}[/dispmath]

A evo i drugog...
[dispmath]\lim_{x\to0}\frac{1-\cos^3x}{x^2}[/dispmath]
Brojilac rastavljamo kao razliku kubova, prema formuli [inlmath]a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)[/inlmath]:
[dispmath]\lim_{x\to0}\frac{\left(1-\cos x\right)\left(1+\cos x+\cos^2x\right)}{x^2}=\lim_{x\to0}\left(1+\cos x+\cos^2x\right)\cdot\lim_{x\to0}\frac{1-\cos x}{x^2}=[/dispmath][dispmath]=\left(1+1+1^2\right)\cdot\lim_{x\to0}\frac{2\sin^2\frac{x}{2}}{4\left(\frac{x}{2}\right)^2}=\frac{3}{2}\lim_{x\to0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2=\frac{3}{2}\cdot1=\frac{3}{2}[/dispmath]