I have two questions for the maths geniuses here! They're both from Q.15 of Sheet 1 in the Calculus 2 problem booklet.
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(Sorry for the giant picture)
How do? D:
I can't work out how any of the parts would give me something finite for both ends of [value]<[expression]<[value].
For part a, you can use 3^n+1>3^n but 3^n+1<2*3^n
So, we have 3^n<3^n + 1 < 2*3^n
Raising all three of these to the power of 1/n:
3<(3^n+1)^n<2^(1/n)*3
As n approaches infinity, 1/n approaches 0 so 2^(1/n) approaches 1. Thus the first and third limits approach 3 (well the first one IS 3 identically), so by the Sandwich Theorem your limit is 3. Sort of makes sense if you see how irrelevant the +1 becomes in the bracket.
Of course, you could also have done this by taking out a factor of 3^n: (3^n + 1)^(1/n) = 3*(1 + 3^-n)^(1/n). The power approaches zero and the thing inside the bracket approaches 1, leaving the constant 3 outside.
As for b...I think you can just tell that this will be 0. Oh well, time for a better proof.
In all honesty, I don't know why you'd need the Sandwich theorem for this one. It's so obvious if you write out each individual term (you know, how n!/n^n = 1/n^n * 2/n^n * 3/n^n etc). Oh well.
You can appreciate that n!/n^n is smaller than 1/n as if you write out all of the terms above and replace all of the terms in the numerator except 1 with n, you get 1/n. Similarly, n!/n^n is clearly larger than 0.
So we have 0<n!/n^n<1/n. What happens if we let n become large? The third term becomes zero, the first term IS zero and the second term is stuck between. GG limit question.
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