I used L'hospitals to answer this question and got 0. The back of the book said it was 1. Anyone care to walk me through this problem please.
Here goes:-
lim_(t->∞) (t - 2) / (√(t² - 5) - 3)
The limit of a sum is the sum of the limits:
= [lim_(t->∞) (t)/(√(t² - 5) - 3)] – [2 (lim_(t->∞)/(√(t² - 5) - 3))]
Indeterminate form of type∞/∞.
Using L'Hospital's rule we have,
lim_(t->∞) (t)/(√(t² - 5) - 3) = lim_(t->∞) (( dt)/( dt))/(( d(√(t² - 5) - 3))/( dt)):
=[ lim_(t->∞) √(t² - 5)/t] – [2/(lim_(t->∞) (√(t² - 5) - 3))]
Simplify radicals,
√t(t² - 5)/t = √((t² - 5)/t²):
= [lim_(t->∞) √((t² - 5) / t²)] – [2/(lim_(t->∞) (√(t² - 5) - 3))]
Using the power law, write lim_(t->∞) √((t²-5)/t²) as √(lim_(t->∞) (t²-5)/t²
=[ √t(lim_(t->∞) (t² - 5)/t²)] – [2/(lim_(t->∞) (√(t² - 5) - 3))]
Indeterminate form of type ∞/∞.
lim_(t->∞) (t² - 5)/t² = lim_(t->∞) (( d(t² - 5)) / ( dt))/(( dt²)/( dt)):
The limit of a constant is the constant:
= 1 – [2/(lim_(t->∞) (√(t² - 5) - 3))]
= 1 – [2/(lim_(t->∞) √(t² - 5) - 3)]
Using the power law, write lim_(t->∞) √(t² - 5) as √(lim_(t->∞) (t² - 5)):
= 1 – [2/(√(lim_(t->∞) (t² - 5)) - 3)]
The limit of t² - 5 as t approaches ∞ is∞:
1 - 0
Answer: = 1
http://tutorial.math.lamar.edu/Classes/CalcI/LHosp...
MY EYES!!!!
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Answers & Comments
Verified answer
Here goes:-
lim_(t->∞) (t - 2) / (√(t² - 5) - 3)
The limit of a sum is the sum of the limits:
= [lim_(t->∞) (t)/(√(t² - 5) - 3)] – [2 (lim_(t->∞)/(√(t² - 5) - 3))]
Indeterminate form of type∞/∞.
Using L'Hospital's rule we have,
lim_(t->∞) (t)/(√(t² - 5) - 3) = lim_(t->∞) (( dt)/( dt))/(( d(√(t² - 5) - 3))/( dt)):
=[ lim_(t->∞) √(t² - 5)/t] – [2/(lim_(t->∞) (√(t² - 5) - 3))]
Simplify radicals,
√t(t² - 5)/t = √((t² - 5)/t²):
= [lim_(t->∞) √((t² - 5) / t²)] – [2/(lim_(t->∞) (√(t² - 5) - 3))]
Using the power law, write lim_(t->∞) √((t²-5)/t²) as √(lim_(t->∞) (t²-5)/t²
=[ √t(lim_(t->∞) (t² - 5)/t²)] – [2/(lim_(t->∞) (√(t² - 5) - 3))]
Indeterminate form of type ∞/∞.
Using L'Hospital's rule we have,
lim_(t->∞) (t² - 5)/t² = lim_(t->∞) (( d(t² - 5)) / ( dt))/(( dt²)/( dt)):
The limit of a constant is the constant:
= 1 – [2/(lim_(t->∞) (√(t² - 5) - 3))]
The limit of a constant is the constant:
The limit of a sum is the sum of the limits:
= 1 – [2/(lim_(t->∞) √(t² - 5) - 3)]
Using the power law, write lim_(t->∞) √(t² - 5) as √(lim_(t->∞) (t² - 5)):
= 1 – [2/(√(lim_(t->∞) (t² - 5)) - 3)]
The limit of t² - 5 as t approaches ∞ is∞:
1 - 0
Answer: = 1
http://tutorial.math.lamar.edu/Classes/CalcI/LHosp...
MY EYES!!!!