求 lim(x→∞)[sin(2/x)+cos(1/x)]^x的极限.
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求 lim(x→∞)[sin(2/x)+cos(1/x)]^x的极限.
lim(x→∞)[sin(2/x)+cos(1/x)]^x的极限.需要详细步骤.
lim(x→∞)[sin(2/x)+cos(1/x)]^x
=lim(x→∞)[1+(sin(2/x)+cos(1/x)-1)]^x
=lim(x→∞)[1+(sin(2/x)+cos(1/x)-1)]^[1/((sin(2/x)+cos(1/x)-1))]*[x*((sin(2/x)+cos(1/x)-1))]
=e^lim[x*((sin(2/x)+cos(1/x)-1))]
=e^lim(t→0)[((sin(2t)+cos(t)-1))/t]
=e^lim(t→0)[2(sin(2t)/2t+(cos(t)-1)/t)]
=e^(2-0)=e^2
lim(x→∞)[sin(2/x)+cos(1/x)]^x
=lim(x→∞)[1+(sin(2/x)+cos(1/x)-1)]^x
=lim(x→∞)[1+(sin(2/x)+cos(1/x)-1)]^[1/((sin(2/x)+cos(1/x)-1))]*[x*((sin(2/x)+cos(1/x)-1))]
=e^lim[x*((sin(2/x)+cos(1/x)-1))]
=e^lim(t→0)[((sin(2t)+cos(t)-1))/t]
=e^lim(t→0)[2(sin(2t)/2t+(cos(t)-1)/t)]
=e^(2-0)=e^2
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