f(x) = \([\frac{min(t^2+4t+6)sin x}x]\)
∵ t2 + 4t + 6 = (t + 2)2 + 2 \(\geq\) 2
∴ min (t2 + 4t + 6) = 2
∴ f(x) = \([\frac{2sinx}x]\)
∵ sin x \(\leq\) x
⇒ -1 < \(\frac{sin x}x\leq1\)
⇒ -2 < \(\frac{2sin x}x\leq2\)
∴ \([\frac{2sinx}x]\) = \(\begin{cases}2&;x=0or sinx/x=1\\1&;1/2 \leq\frac{sin x}x<1\\0&;0\leq\frac{sin x}x<1/2\\-1&;\frac{sin x}x<0\end{cases}\)
\(\lim\limits_{x\to 0}f(x) = \lim\limits_{x\to 0}[\frac{2sin x}x]\) = 1