Question

Tìm giới hạn của hàm số sau:

\(\lim\limits_{x\rightarrow1^+}\left(x-1\right)\sqrt{\dfrac{2x+3}{x^2-1}}\)

Answer 1

\(\left(x-1\right)\sqrt{\dfrac{2x+3}{x^2-1}}=\sqrt{\dfrac{\left(x-1\right)\left(2x+3\right)}{x+1}}=\sqrt{2x-2+\dfrac{x-1}{x+1}}\)

Ta có:

\(\lim\limits_{x\rightarrow1^+}\left(x-1\right)\sqrt{\dfrac{2x+3}{x^2-1}}=\lim\limits_{x\rightarrow1^+}\sqrt{2x-2+\dfrac{x-1}{x+1}}=\sqrt{2-2+\dfrac{1-1}{1+1}}=0\)

2x-2 > 0 với mọi x>1

\(\dfrac{x-1}{x+1}\)>0 với mọi x>1

=>\(\lim\limits_{x\rightarrow1^+}\left(x-1\right)\sqrt{\dfrac{2x+3}{x^2-1}}=+\infty\)