求极限limh→0{[1/(5+h)]-1/5}/h1.limh→0{[1/(5+h)]-1/5}/h2.limx→2[√(x+2)-√(3x-2)]/(√5x-1)-(√4x+1)3.limh→0{[³√(1+h)]-1}/h

Question

求极限 lim h →0 {[1/(5+h)]-1/5}/h
1.lim h →0 {[1/(5+h)]-1/5}/h
2.lim x→2 [√(x+2)-√(3x-2)]/ (√5x-1)-(√4x+1)
3.lim h →0 {[³√(1+h)]-1}/h

Answer 1

1.lim h →0 {[1/(5+h)]-1/5}/h
=lim h →0 { (5)/[ 5(5+h) ] - (5-h)/ [ 5(5+h) ] }/h
= lim h →0 { (5-5-h)/[ 5(5+h) ] }/h
= lim h →0 -h/ [ 5h(5+h) ]
= -1/25
2.lim x→2 [√(x+2)-√(3x-2)]/ √(5x-1)-√(4x+1)
= lim x→2 { (x+2-3x+2)[ (√(5x-1)+√(4x+1)) ]/ (5x-1+4x-1)[√(x+2)-√(3x-2) ] }
=lim x→2 { -2(X-2)[ (√(5x-1)+√(4x+1)) ]/ X-2)[√(x+2)-√(3x-2) ] }
=-3
3.lim h →0 {[³√(1+h)]-1}/h
=lim h →0 {[³√(1+h)]-1}{³√(1+h)^2 +³√(h+1)+1 }/h{³√(1+h)^2 +³√(h+1)+1 }
=lim h →0 {[h+1-1}/h{³√(1+h)^2 +³√(h+1)+1 }
=1/(1+1+1)
=1/3