Find the limit of [1/(t x (1+t)^(1/2)) - (1/t)] as x approaches zero.
The answer is -1/2 according to my book, but I can't get a rational answer. Please help.
lim t-->0 [1/(t x (1+t)^0.5) - (1/t)]
= lim t --> [(1-(1+t)^0.5)/(t(1+t)^0.5)]
= lim t -->0 [(1+t)^0.5 - 1 - t)/(t + t^2)
Now, use L'Hopital's rule to get:
= lim t -->0 [{1/(2(t+1)^0.5) - 1} / (1+2t)
Plug 0 in for t to get
= (1/2 - 1)/1
= -1/2
I hope this helps!
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Verified answer
lim t-->0 [1/(t x (1+t)^0.5) - (1/t)]
= lim t --> [(1-(1+t)^0.5)/(t(1+t)^0.5)]
= lim t -->0 [(1+t)^0.5 - 1 - t)/(t + t^2)
Now, use L'Hopital's rule to get:
= lim t -->0 [{1/(2(t+1)^0.5) - 1} / (1+2t)
Plug 0 in for t to get
= (1/2 - 1)/1
= -1/2
I hope this helps!