Subject 1 Determine the real numbers a and b such that: $\lim _{x \rightarrow 2} \frac{\sqrt{x^{2}-x+a}-b}{x-2}=\frac{3}{4}$.
## Solution and Grading

If $\sqrt{2+a}-b>0, \lim _{x \nearrow 2} \frac{\sqrt{x^{2}-x+a}-b}{x-2}=-\infty$ and $\lim _{x \gg 2} \frac{\sqrt{x^{2}-x+a}-b}{x-2}=\infty$.

The given limit does not exist, contradiction. Similarly for $\sqrt{2+a}-b<0$. It follows that $\sqrt{2+a}-b=0$.

For $b=\sqrt{2+a}$,

$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{\sqrt{x^{2}-x+a}-\sqrt{2+a}}{x-2}=\lim _{x \rightarrow 2} \frac{x^{2}-x-2}{(x-2)\left(\sqrt{x^{2}-x+a}+\sqrt{2+a}\right)}= \\
& \lim _{x \rightarrow 2} \frac{x+1}{\sqrt{x^{2}-x+a}+\sqrt{2+a}}=\frac{3}{2 \sqrt{2+a}}
\end{aligned}
$$

Thus $\sqrt{2+a}=2 \rightarrow a=2$. In conclusion, $a=2, b=2$.

$\sqrt{2+a}-b \neq 0$, impossible. $.3 p$

$\sqrt{2+a}-b=0, \lim _{x \rightarrow 2} \frac{\sqrt{x^{2}-x+a}-b}{x-2}=\frac{3}{2 \sqrt{2+a}}$

$\sqrt{2+a}=2 \rightarrow a=2$. $.1 \mathrm{p}$

$b=\sqrt{2+a}=2$ $1 \mathrm{p}$