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how to solve this two question?
Date Posted: 5 years ago
Views: 265
As h→ ∞, 1/h → 0
So (1 + h) → 1
And thus (1 + h)^h → 1
for the 2nd one, apply L'Hopital's rule. Differentiate both numerator and denominator.
So lim (a^h -1)/h = lim a^h ln(a) / 1
= a^h ln (a)
As h→0, a^h → 1
So a^h ln (a) → ln (a)
Now, since a →e, ln (a) → ln (e) = 1
So the limit is 1
So (1 + h) → 1
And thus (1 + h)^h → 1
for the 2nd one, apply L'Hopital's rule. Differentiate both numerator and denominator.
So lim (a^h -1)/h = lim a^h ln(a) / 1
= a^h ln (a)
As h→0, a^h → 1
So a^h ln (a) → ln (a)
Now, since a →e, ln (a) → ln (e) = 1
So the limit is 1
Not so straightforward. I am very sure the limit for the first one is e, not 1.
True. thanks for pointing it out
thanks eric
See 2 Answers
done
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First question; this is a well known result and in fact this was how the constant e was discovered many years ago.
Date Posted:
5 years ago
done
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Second question.
L' Hopital's rule can be invoked when the value of
Numerator/Denominator
is either 0/0 or infinity/infinity.
If the first application of the L' Hopital's rule still results in either 0/0 or infinity/infinity, we must do another application of the L' Hopital rule. This is to be done until the limit is no longer 0/0 or infinity/infinity.
L' Hopital's rule can be invoked when the value of
Numerator/Denominator
is either 0/0 or infinity/infinity.
If the first application of the L' Hopital's rule still results in either 0/0 or infinity/infinity, we must do another application of the L' Hopital rule. This is to be done until the limit is no longer 0/0 or infinity/infinity.
Date Posted:
5 years ago
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