>>267
(1)
{ √( 2n^2 -n) } / { √(n^2 +1) + n}
= { √(2 - (1/n)) } / {√(1+(1/n^2)) + 1} → (√2)/2 (n→∞)

(2) 2/[n-√{n^(2)+3n}]
= 2{ n + √( n^2 +3n) } / { n^2 - (n^2 +3n)}
= 2{ n + √( n^2 +3n) } /(-3n)
= - (2/3) { 1 + √(1+(3/n))} → -(2/3) { 1+ 1} = -4/3 (n→∞)

(3) n^2 -2n = (n-1)^2 -1 → ∞ (n→∞)


(4) 5n-3n^2 = - 3(n^2 - (5/3)n) = -3 { n-(5/6)}^2 + (25/12) → -∞ (n→∞)

1/{(2k-1) (2k+1)} = (1/2) ( { 1/(2k-1)} - {1/(2k+1)} )
1/(1*3) = (1/2) ( (1/1) - (1/3) )
1/(3*5) = (1/2) ( (1/3) - (1/5) )
1/(5*7) = (1/2) ( (1/5) - (1/7) )
…
1/{(2n-1) (2n+1)} = (1/2) ( { 1/(2n-1)} - {1/(2n+1)} )

なので、全部足しあわせれば

Σ 1/{(2k-1) (2k+1)} = (1/2) - (1/2) ( 1/(2n+1) ) → (1/2) (n→∞)