求x*sin[2x/(x^2+1)]的极限
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求x*sin[2x/(x^2+1)]的极限
x是趋向于什么?不然没法求极限的
再问: 不好意思,忘了写了,x趋向于无穷大。
再答: x→∞时2x/(x^2+1)→0 所以lim(x→∞)sin[2x/(x^2+1)]/[2x/(x^2+1)]=1(等价于x→0时limsinx/x=1) lim(x→∞)2x^2/(x^2+1)=lim2/(1+1/x^2)=2 所以lim(x→∞)x*sin[2x/(x^2+1)] =lim(x→∞){sin[2x/(x^2+1)]/[2x/(x^2+1)]}*[2x^2/(x^2+1)] =1*2 =2
再问: 不好意思,忘了写了,x趋向于无穷大。
再答: x→∞时2x/(x^2+1)→0 所以lim(x→∞)sin[2x/(x^2+1)]/[2x/(x^2+1)]=1(等价于x→0时limsinx/x=1) lim(x→∞)2x^2/(x^2+1)=lim2/(1+1/x^2)=2 所以lim(x→∞)x*sin[2x/(x^2+1)] =lim(x→∞){sin[2x/(x^2+1)]/[2x/(x^2+1)]}*[2x^2/(x^2+1)] =1*2 =2
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