图中打钩的题目怎么做啊,求解答啊。。。
1个回答
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(2)lim(x->1) (x^2+3x+1)/(x-1)^2=1/0=∞
(4)lim(x->∞) (x-sinx)/(x+sinx)=(1-sinx/x)/(1+sinx/x)=1/1=1
(6)lim(x->∞) [2^(n+1)+3^(n+1)]/[2^n+3^n]
=lim(x->∞) [2*(2/3)^n+3]/[(2/3)^n+1]
=3/1=3
(8)lim(x->0) (1-3x)^(1/x+1)
=e^{lim(x->0) ln[(1-3x)^(1/x+1)]}
=e^{lim(x->0) (1/x+1)*ln(1-3x)}
=e^{lim(x->0) ln(1-3x)/[x/(x+1)]}
=e^{lim(x->0) [-3/(1-3x)]/[1/(x+1)^2]} 洛必达法则
=e^{lim(x->0) [-3(x+1)^2/(1-3x)]}
=e^{-3/1}
=1/e^3
(4)lim(x->∞) (x-sinx)/(x+sinx)=(1-sinx/x)/(1+sinx/x)=1/1=1
(6)lim(x->∞) [2^(n+1)+3^(n+1)]/[2^n+3^n]
=lim(x->∞) [2*(2/3)^n+3]/[(2/3)^n+1]
=3/1=3
(8)lim(x->0) (1-3x)^(1/x+1)
=e^{lim(x->0) ln[(1-3x)^(1/x+1)]}
=e^{lim(x->0) (1/x+1)*ln(1-3x)}
=e^{lim(x->0) ln(1-3x)/[x/(x+1)]}
=e^{lim(x->0) [-3/(1-3x)]/[1/(x+1)^2]} 洛必达法则
=e^{lim(x->0) [-3(x+1)^2/(1-3x)]}
=e^{-3/1}
=1/e^3
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