Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ và $\lim\limits_{x\to1}\dfrac{f(x)-3}{x^2-x}=2$. Tính $T=\lim\limits_{x\to1}\dfrac{f^2(x)+f(x)-12}{x^2+6x-7}$.
 | $P=\dfrac{9}{4}$ |
 | $P=\dfrac{13}{4}$ |
 | $T=\dfrac{5}{4}$ |
 | $T=\dfrac{7}{4}$ |
Chọn phương án D.
Vì $\lim\limits_{x\to1}\dfrac{f(x)-3}{x^2-x}=\lim\limits_{x\to1}\dfrac{f(x)-3}{x(x-1)}$ hữu hạn nên tồn tại hàm số $g(x)$ sao cho $f(x)-3=(x-1)g(x)$, hay $f(x)=3+(x-1)g(x)$.
Khi đó $\begin{aligned}[t]
\lim\limits_{x\to1}\dfrac{f(x)-3}{x(x-1)}&=\lim\limits_{x\to1}\dfrac{(x-1)g(x)}{x(x-1)}\\
&=\lim\limits_{x\to1}\dfrac{g(x)}{x}=\dfrac{\lim\limits_{x\to1}g(x)}{\lim\limits_{x\to1}x}=\lim\limits_{x\to1}g(x).
\end{aligned}$
Theo đó $\lim\limits_{x\to1}f(x)=\lim\limits_{x\to1}\big(3+(x-1)g(x)\big)=3$.
Vậy $\begin{aligned}[t]
T&=\lim\limits_{x\to1}\dfrac{f^2(x)+f(x)-12}{x^2+6x-7}\\
&=\lim\limits_{x\to1}\dfrac{\big(f(x)+4\big)\big(f(x)-3\big)}{(x-1)(x+7)}\\
&=\lim\limits_{x\to1}\dfrac{\big(f(x)+4\big)(x-1)g(x)}{(x-1)(x+7)}\\
&=\lim\limits_{x\to1}\dfrac{\big(f(x)+4\big)g(x)}{x+7}\\
&=\dfrac{(3+4)\cdot2}{1+7}=\dfrac{7}{4}.
\end{aligned}$