Question

高数（极限与连续）

高数（极限与连续）期盼着大神能帮我解决（步骤详细）😊

Answer 1

令t=x+8，则t=0
原式=[sqrt(9-t)-3]/[2+(t-8)^(1/3)]
=3/2*[sqrt(1-t/9)-1]/[1-(1-t/8)^(1/3)]
使用无穷小替换
=3/2*(-t/18)/(t/24)
=-3/2*24/18=-3/2*4/3=-2

Answer 2

lim(x->-8) [√(1-x) -3 ]/[ 2+x^(1/3) ]
=lim(x->-8) [(1-x) -9 ]/{ [ 2+x^(1/3) ] . [√(1-x) +3 ] }
=lim(x->-8) (-8-x)/{ [ 2+x^(1/3) ] . [√(1-x) +3 ] }
=(1/6) lim(x->-8) (-8-x)/ [ 2+x^(1/3) ]
=-(1/6) lim(x->-8) (8+x)/ [ 2+x^(1/3) ]
=-(1/6) lim(x->-8) [2+x^(1/3)]. [ 4-2x^(1/3)+ x^(2/3) ]/ [ 2+x^(1/3) ]
=-(1/6) lim(x->-8) [ 4-2x^(1/3)+ x^(2/3) ]
=-(1/6) [ 4 +4 +4]
=-2