求大神帮解一下求极限的那四道题 10
1个回答
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(1)原式=lim(x->1) [x-e^(xlnx)]/(1-x+lnx)
=lim(x->1) [1-x^x*(lnx+1)]/(1/x-1)
=lim(x->1) [x^(x+1)*(lnx+1)-x]/(x-1)
=lim(x->1) [x^x+(lnx+1)*x^(x+1)*(lnx+1+1/x)-1]
=2
(2)原式=lim(x->0) [x-ln(1+x)]/xln(1+x)
=lim(x->0) [x-ln(1+x)]/x^2
=lim(x->0) [1-1/(1+x)]/2x
=lim(x->0) 1/2(1+x)
=1/2
(3)原式=lim(x->+∞) e^{x*[ln(2/π)+ln(arctanx)]}
=lim(x->+∞) e^{[1/(1+x^2)arctanx]/(-1/x^2)}
=lim(x->+∞) e^[-x^2/(1+x^2)arctanx]
=lim(x->+∞) e^[-2x/(2x*arctanx+1)]
=lim(x->+∞) e^[-1/(arctanx+1/2x)]
=e^0
=1
(4)令t=1/x
原式=lim(t->0) [(a1^t+a2^t+...+an^t)/n]^(n/t)
=lim(t->0) e^{(n/t)*[ln(a1^t+a2^t+...+an^t)/n]}
=lim(t->0) e^[n*(lna1*a1^t+lna2*a2^t+...+lnan*an^t)/(a1^t+a2^t+...+an^t)]
=e^(lna1+lna2+...+lnan)
=a1*a2*...*an
=lim(x->1) [1-x^x*(lnx+1)]/(1/x-1)
=lim(x->1) [x^(x+1)*(lnx+1)-x]/(x-1)
=lim(x->1) [x^x+(lnx+1)*x^(x+1)*(lnx+1+1/x)-1]
=2
(2)原式=lim(x->0) [x-ln(1+x)]/xln(1+x)
=lim(x->0) [x-ln(1+x)]/x^2
=lim(x->0) [1-1/(1+x)]/2x
=lim(x->0) 1/2(1+x)
=1/2
(3)原式=lim(x->+∞) e^{x*[ln(2/π)+ln(arctanx)]}
=lim(x->+∞) e^{[1/(1+x^2)arctanx]/(-1/x^2)}
=lim(x->+∞) e^[-x^2/(1+x^2)arctanx]
=lim(x->+∞) e^[-2x/(2x*arctanx+1)]
=lim(x->+∞) e^[-1/(arctanx+1/2x)]
=e^0
=1
(4)令t=1/x
原式=lim(t->0) [(a1^t+a2^t+...+an^t)/n]^(n/t)
=lim(t->0) e^{(n/t)*[ln(a1^t+a2^t+...+an^t)/n]}
=lim(t->0) e^[n*(lna1*a1^t+lna2*a2^t+...+lnan*an^t)/(a1^t+a2^t+...+an^t)]
=e^(lna1+lna2+...+lnan)
=a1*a2*...*an
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