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