6. Given $f(x)=x^{3}$ on $[1, b]$ satisfies
$$
\frac{f(b)-f(1)}{b-1}=f^{\prime}(t)(1<t<b) \text {. }
$$

then $\lim _{b \rightarrow 1} \frac{t-1}{b-1}=$
6. $\frac{1}{2}$.
$$
\begin{array}{l}
\lim _{b \rightarrow 1} \frac{t-1}{b-1}=\lim _{b \rightarrow 1} \frac{\sqrt{\frac{1}{3}\left(b^{2}+b+1\right)}-1}{b-1} \\
=\lim _{b \rightarrow 1} \frac{\sqrt{b^{2}+b+1}-\sqrt{3}}{\sqrt{3}(b-1)} \\
=\lim _{b \rightarrow 1} \frac{(b-1)(b+2)}{\sqrt{3}(b-1)\left(\sqrt{b^{2}+b+1}+\sqrt{3}\right)} \\
=\lim _{b \rightarrow 1} \frac{b+2}{\sqrt{3}\left(\sqrt{b^{2}+b+1}+\sqrt{3}\right)} \\
=\frac{1+2}{\sqrt{3}\left(\sqrt{1^{2}+1+1}+\sqrt{3}\right)}=\frac{1}{2} .
\end{array}
$$

Explanation: The proposition has a background in the Mean Value Theorem for derivatives. If $\frac{f(b)-f(a)}{b-a}=f^{\prime}(\xi)$, then $\lim _{b \rightarrow a} \frac{\xi-a}{b-a}=\frac{1}{2}$.