Question

\(\lim\limits_{x\rightarrow2}\left(\dfrac{\sqrt{mx+n}-1}{x^2-4}\right)=1\)

Tính tổng m+n

Answer 1

0

Answer 2

Do giới hạn đã cho hữu hạn \(\Rightarrow\sqrt{mx+n}-1=0\) có nghiệm \(x=2\)

\(\Rightarrow\sqrt{2m+n}=1\Rightarrow n=-2m+1\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{mx-2m+1}-1}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{mx-2m}{\left(x-2\right)\left(x+2\right)\left(\sqrt{mx-2m+1}+1\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{m}{\left(x+2\right)\left(\sqrt{mx-2m+1}+1\right)}=\dfrac{m}{4\left(\sqrt{2m-2m+1}+1\right)}=\dfrac{m}{8}\)

\(\Rightarrow\dfrac{m}{8}=1\Rightarrow m=8\Rightarrow n=-15\)