Question

Tính giới hạn: \(limn\left(1-\dfrac{n^2+3}{n^2-1}\right)\)

Answer 1

Lời giải:
\(\lim\limits (1-\frac{n^2+3}{n^2-1})=\lim \frac{-4}{n^2-1}=0\) do $n^2-1\to \infty$ khi $n\to \infty$

Answer 2

\(\lim\limits\left[n\left(1-\dfrac{n^2+3}{n^2-1}\right)\right]\)

\(=\lim\limits\left[n\cdot\dfrac{n^2-1-n^2-3}{n^2-1}\right]\)

\(=\lim\limits\dfrac{-4n}{n^2-1}\)

\(=\lim\limits\dfrac{-\dfrac{4}{n}}{1-\dfrac{1}{n^2}}=\dfrac{0}{1-0}=0\)