Question

tìm a,b sao cho

\(\lim_{x\rightarrow2}\dfrac{x^{2}+2ax-b}{x^{2}-4}=4\)

 

 

Answer 1

Giới hạn đã cho hữu hạn nên \(x^2+2ax-b=0\) có nghiệm \(x=2\)

\(\Rightarrow4+4a-b=0\Rightarrow b=4a+4\)

\(\Rightarrow\lim\limits_{x\rightarrow2}\dfrac{x^2+2ax-4a-4}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2a+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x+2a+2}{x+2}=\dfrac{2a+4}{4}=4\)

\(\Rightarrow a=6\Rightarrow b=28\)