Najd�te $a,$ $b\in\R,$ aby $\lnul\frac{x - (a + b\cos x)\sin x}{x^4} = 0$.
\reseni Spo�teme limitu pomoc� Taylorova vzorce:
\begin{multline*}
    \lnul\frac{x - (a + b\cos x)\sin x}{x^4} =\\
   =\lnul\frac{x - \left(a + b - \frac{bx}2 + \frac{bx^2}{24} + b\cdot o(x^4)\right)\left(x-\frac{x^3}6 + o(x^4)\right)}{x^4}=\\
   =\lnul\frac{x - \left(x(a+b) + x^3\left(-\frac a6 -\frac{2b}3\right) + o(x^4)\right)}{x^4}=\\
   =\lnul\frac{x(1 - a - b) + x^3\left(\frac a6 + \frac{2b}3\right) + o(x^4)}{x^4}.
\end{multline*}
Tato limita se rovn� nule pr�v� tehdy, kdy� koeficienty u $x$ a $x^3$ jsou rovny nule. M�me tedy �e�it
n�sleduj�c� soustavu rovnic:
%V TeXu nen� dobr� zp�sob pro zobrazen� soustavy rovnic, proto mus�m pou�it ten nep�ehledn�:
\begin{align*}
    &\left\lbrace\begin{aligned}
        1-a-b&=0\\
        a + 4b&=0
    \end{aligned}\right.\\
    &\left\lbrace\begin{aligned}
        a &= -4b\\
        1+3b&=0
    \end{aligned}\right.\\
    &\left\lbrace\begin{aligned}
        b&=-\frac13\\
        a&=-\frac43.
    \end{aligned}\right.
\end{align*}
