如何求x→+∞时的函数极限?
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x→+∞limx²[arctan(x+1)-arctanx]【∞•0】
=x→+∞lim[arctan(x+1)-arctanx]/(1/x²)【0/0】
=x→+∞lim{1/[1+(x+1)²]-1/(1+x²)}/(-2/x³)
=x→+∞lim{[1/(x²+2x+2)-1/(1+x²)]/(-2/x³)}
=x→+∞lim{[(1+x²)-(x²+2x+2)]/(x²+2x+2)(1+x²)}/(-2/x³)
=x→+∞lim[(-2x-1)/(x^4+2x³+3x²+2x+2)]/(-2/x³)
=(1/2)x→+∞lim[(2x^4+x³)/(x^4+2x³+3x²+2x+2)]
=(1/2)x→+∞lim[2+(1/x)]/[1+(2/x)+(3/x²)+(2/x³)+(2/x^4)]
=(1/2)×2=1;
=x→+∞lim[arctan(x+1)-arctanx]/(1/x²)【0/0】
=x→+∞lim{1/[1+(x+1)²]-1/(1+x²)}/(-2/x³)
=x→+∞lim{[1/(x²+2x+2)-1/(1+x²)]/(-2/x³)}
=x→+∞lim{[(1+x²)-(x²+2x+2)]/(x²+2x+2)(1+x²)}/(-2/x³)
=x→+∞lim[(-2x-1)/(x^4+2x³+3x²+2x+2)]/(-2/x³)
=(1/2)x→+∞lim[(2x^4+x³)/(x^4+2x³+3x²+2x+2)]
=(1/2)x→+∞lim[2+(1/x)]/[1+(2/x)+(3/x²)+(2/x³)+(2/x^4)]
=(1/2)×2=1;