Question

求下列极限，过程

Answer 1

(1)
lim(x->π/4) 2(sin2x)^2
=2(1)^2
=2
(2)
lim(x->1)[(x^2-1)/(tanx.cosx)]^7
=[ 0/(tan1.cos1)]^7
=0
(3)
lim(x->2) ln(x-1)cos2x
=(ln1).cos4

=0
(4)

lim(x->0) [√(4x+4) -2 ]/x
=lim(x->0) [(4x+4) -4 ]/{ x.[√(4x+4) +2 ]}
=lim(x->0) 4x/{ x.[√(4x+4) +2 ]}
=lim(x->0) 4/[√(4x+4) +2 ]
=4/(2+2)
=1
(5)
lim(x->+∞) [√(x^2+3x)-√(x^2-x)]
=lim(x->+∞) [(x^2+3x)-(x^2-x)]/[√(x^2+3x)+√(x^2-x)]
=lim(x->+∞) 4x/[√(x^2+3x)+√(x^2-x)]
=lim(x->+∞) 4/[√(1+3/x^2)+√(1-1/x)]
=4/(1+1)
=2
(6)
lim(x->0) [√(1+5x)-√(1-3x)]/x
=lim(x->0) [(1+5x)-(1-3x)]/{ x.[√(1+5x)+√(1-3x)] }
=lim(x->0) 8x/{ x.[√(1+5x)+√(1-3x)] }
=lim(x->0) 8/[√(1+5x)+√(1-3x)]
=8/(1+1)
=4