Can anyone help me solve this using L'Hospital's rule
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Can anyone help me solve this using L'Hospital's rule

[From: ] [author: ] [Date: 12-11-19] [Hit: ]
If you do,You get 0/0, which means we can use the rule.Differentiate the top,Differentiate the bottom,If you do,......
limit as x approaches 0 (1/t-(1/(t^2+t)).

The answer is supposed to be 1 but how did they get that? Please give a good explanation.
Thanks in advance.

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1/t - 1/(t^2 + t)

Combine the fractions because L'Hospital's rule works for infinity/infinity or 0/0

If you do, you get

t/(t^2+t)

Apply the limit at t goes to 0

You get 0/0, which means we can use the rule.

Differentiate the top, you get 1
Differentiate the bottom, you get 2t+1

Combine so you apply the limit as t goes to 0 for 1/(2t+1)

If you do, the 2t goes away, and you're left with 1/1 = 1

Done.

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Combine the fractions to turn the indeterminate form ∞-∞ into ∞/∞, then apply L'Hôpital's directly.

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(1/t-(1/(t^2+t)) = [t^2+t-t]/[t(t^2+t)]=
t^2/[t^3+t^2]
Apply L'Hopital twice then
lim t-->0 2/[6t+2] = 2/2 =1
1
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