Giả sử \(f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d\)\(\left( a\ne 0 \right)\).

Có \(f'\left( x \right) = 3a{x^2} + 2bx + c = 0 \Leftrightarrow \left[ \begin{array}{l} x = {x_1}\\ x = {x_2} = {x_1} + 2 \end{array} \right.\).

Suy ra: \({f}'\left( x \right)=3a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\)

\(\Rightarrow {f}'\left( x \right)=3a\left( x-{{x}_{1}} \right)\left( x-{{x}_{1}}-2 \right)\)

\(\Rightarrow {f}'\left( x \right)=3a{{\left( x-{{x}_{1}} \right)}^{2}}-6a\left( x-{{x}_{1}} \right)\).

Lấy nguyên hàm hai vế ta có:

\(f\left( x \right)=a{{\left( x-{{x}_{1}} \right)}^{3}}-3a{{\left( x-{{x}_{1}} \right)}^{2}}+C\).

Khi đó \(f\left( {{x}_{1}} \right)=C\) và \(\,\,\,\,f\left( {{x}_{2}} \right)=a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{3}}-3a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+C=8a-12a+C=C-4a\).

Mà \(f\left( {{x}_{1}} \right)+\,\,f\left( {{x}_{2}} \right)=0\), nên \(C+C-4a=0\)\(\Leftrightarrow C=2a\).

Suy ra \(f\left( x \right)=a{{\left( x-{{x}_{1}} \right)}^{3}}-3a{{\left( x-{{x}_{1}} \right)}^{2}}+2a\).

Mặt khác \(\int\limits_{{{x}_{1}}}^{{{x}_{1}}+1}{f\left( x \right)\text{d}x=\frac{5}{4}}\,\,\Leftrightarrow \int\limits_{{{x}_{1}}}^{{{x}_{1}}+1}{\left[ a{{\left( x-{{x}_{1}} \right)}^{3}}-3a{{\left( x-{{x}_{1}} \right)}^{2}}+2a \right]\text{d}x=\frac{5}{4}}\)

\(\Leftrightarrow \left. \left[ \frac{a}{4}{{\left( x-{{x}_{1}} \right)}^{4}}-a{{\left( x-{{x}_{1}} \right)}^{3}}+2ax \right]_{{}}^{{}} \right|_{\,{{x}_{1}}}^{\,{{x}_{1}}+1}=\frac{5}{4}\)\(\Leftrightarrow \left[ \frac{a}{4}-a+2a\left( {{x}_{1}}+1 \right) \right]-2a{{x}_{1}}=\frac{5}{4}\)  \(\Leftrightarrow a=1\).

Do đó: \(f\left( x \right)={{\left( x-{{x}_{1}} \right)}^{3}}-3{{\left( x-{{x}_{1}} \right)}^{2}}+2\).

Vậy \(L=\underset{x\to \,{{x}_{1}}}{\mathop{\lim }}\,\frac{f\left( x \right)-2}{{{\left( x-{{x}_{1}} \right)}^{2}}}=\underset{x\to \,{{x}_{1}}}{\mathop{\lim }}\,\frac{{{\left( x-{{x}_{1}} \right)}^{3}}-3{{\left( x-{{x}_{1}} \right)}^{2}}}{{{\left( x-{{x}_{1}} \right)}^{2}}}=\underset{x\to \,{{x}_{1}}}{\mathop{\lim }}\,\left[ \left( x-{{x}_{1}} \right)-3 \right]=-\,3\).