lim(x->0)f(x)
=lim(x->0)(√(x+4)-2)/x
=lim(x->0)x/[x(√(x+4)+2)] (分子有理化)
=lim(x->0)1/(√(x+4)+2)
=1/(√4+2)
=1/(2+2)
=1/4
因为f(x)在x=0处连续,
所以k=1/4
如仍有疑惑,欢迎追问.
lim(x->0)f(x)
=lim(x->0)(√(x+4)-2)/x
=lim(x->0)x/[x(√(x+4)+2)] (分子有理化)
=lim(x->0)1/(√(x+4)+2)
=1/(√4+2)
=1/(2+2)
=1/4
因为f(x)在x=0处连续,
所以k=1/4
如仍有疑惑,欢迎追问.