Question

当a＞0时，求

lim

x→0

x2+a2 -a

x2+b2 -b

．

Answer 1

原式=

lim

x→0

( x2+a2 -a)( x2+a2 +a)( x2+b2 +b)

( x2+b2 -b)( x2+b2 +b)( x2+a2 +a)

=

lim

x→0

(x2+a2-a2)( x2+b2 +b)

(x2+b2- b 2  )( x2+a2 +a)

=

lim

x→0

x2+b2 +b

x2+a2 +a

=

|b|+b

|a|+a

=

0            (当b≤0时) b a             (当b＞0时).