Question

极限与定积分

设f(t) = ∫[1~(1+1/t)] √(1 + x^t) dx
求lim(t-->∞) [t * f(t)]

Answer 1

好难啊。

Answer 2

f(t)=∫[1, (1+1/t)] √(1+x^t)dx
u=x-1, x=1,u=0, x=1+1/t, u=1/t
=∫[0,1/t] √[1+(u+1)^t] du
lim(t->∝)1/t=0, lim(x->∝) f(t)=0
lim(t->∝)t*f(t)=lim(t->∝)f(t)/(1/t)
=lim(t->∝)f(t)'/(1/t)'

f(t)=g(1/t)=∫[0,1/t]√[1+(u+1)^t]du
dg(1/t)/d(1/t)=√[1+(1/t+1)^t]
f'(t)=dg(1/t)/d(1/t) *(1/t)'=√[1+(1/t+1)^t]*(1/t)'

lim(t->∝)t*f(t)=lim(t->∝)f(t)'/(1/t)'=lim(t->∝)√[1+(1+1/t)^t]=√(1+e)

Answer 3

f(t)=∫[1, (1+1/t)] √(1+x^t)dx
u=x-1, x=1,u=0, x=1+1/t, u=1/t
=∫[0,1/t] √[1+(u+1)^t] du
lim(t->∝)1/t=0, lim(x->∝) f(t)=0
lim(t->∝)t*f(t)=lim(t->∝)f(t)/(1/t)
=lim(t->∝)f(t)'/(1/t)'

f(t)=g(1/t)=∫[0,1/t]√[1+(u+1)^t]du
dg(1/t)/d(1/t)=√[1+(1/t+1)^t]
f'(t)=dg(1/t)/d(1/t) *(1/t)'=√[1+(1/t+1)^t]*(1/t)'

lim(t->∝)t*f(t)=lim(t->∝)f(t)'/(1/t)'=lim(t->∝)√[1+(1+1/t)^t]=√(1+e)

Answer 4

学过
学的不通 也忘完了