回復 45# money 的帖子
第 28 題：
設\(k\)為定數，若\(\displaystyle \lim_{x \to 1}\frac{\sqrt{2x^2+a}-x+b}{(x-1)^2}=k\)，求實數\(a+b+k\)之值= 。(最簡分數)
[解答]
\(\displaystyle \frac{\sqrt{2x^2+a}-x+b}{\left(x-1\right)^2}=\frac{\left(2x^2+a\right)-\left(x-b\right)^2}{\left(x-1\right)^2\left(\sqrt{2x^2+a}+\left(x-b\right)\right)}=\frac{x^2+2bx+a-b^2}{\left(x-1\right)^2\left(\sqrt{2x^2+a}+x-b\right)}\)
因為上列分式多項式當 \(x\to 1\) 時，極限存在，所以 \(\displaystyle (x-1)^2\Bigg|x^2+2bx+a-b^2\Rightarrow \frac{1}{1}=\frac{2b}{-2}=\frac{a-b^2}{1}\Rightarrow a=2,b=-1\)
且 \(\displaystyle k=\lim_{x\to1}\frac{1}{\left(\sqrt{2x^2+a}+x-b\right)}=\lim_{x\to1}\frac{1}{\left(\sqrt{2x^2+2}+x+1\right)}=\frac{1}{4}.\)
\(\displaystyle \Rightarrow a+b+k=\frac{5}{4}\)
