当x→0时,lim[ln(1-2x)+xf(x)]/x^2=4,求lim[f(x-2)]/x

答案：6解法：lim_{x→0}{x[f(x)-2]+2x+ln(1-2x)}/x^2=lim_{x→0}{x[f(x)-2]}/x^2+lim_{x→0}{2x+ln(1-2x)}/x^2=4,
又lim_{x→0}{2x+ln(1-2x)}/x^2=lim_{x→0}{2x+ln(1-2x)}'/[x^2]'=lim_{x→0}{2-2/(1-2x)}/2x=lim_{x→0}
{1-1/(1-2x)}/x=lim_{x→0}{-2/(1-2x)}=-2,所以所求的极限为6
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