[College Calc 01-02. Limits of Sequences] #10

Calculus Level 4

In order to prove that lim ⁡ n → ∞ a n = L , \displaystyle \lim_{n\to\infty}a_n = L, we must find some function N N of ϵ > 0 \epsilon>0 such that for every positive integer n , n, we have

n > N ⟹ ∣ a n − L ∣ < ϵ . n>N \implies |a_n-L|<\epsilon.

Choose all such valid pairs of { a n } \{a_n\} and N . N.

A . a n = log ⁡ n 3 ; N ( ϵ ) = 3 1 / ϵ B . a n = n n 2 + 1 ; N ( ϵ ) = 1 2 ϵ C . a n = n 2 + 2 n − n ; N ( ϵ ) = ( 1 − ϵ ) 2 10 ϵ 2 \begin{aligned} {\rm A.} &&& a_n = \log_n 3; && N(\epsilon) = 3^{1/\epsilon} \\ {\rm B.} &&& a_n = \frac{n}{\sqrt{n^2+1}}; && N(\epsilon) = \frac{1}{\sqrt{2\epsilon}} \\ {\rm C.} &&& a_n = \sqrt{n^2+2n}-n; && N(\epsilon) = \frac{(1-\epsilon)^2}{10\epsilon^2} \end{aligned}

This is a part of the College Calc problem set. You can find more problems here .

None of them A \rm A only B \rm B only C \rm C only A , B \rm A,~B only B , C \rm B,~C only A , C \rm A,~C only All of them

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Let us solve one by one:- We see that :- log n ( 3 ) = 1 log 3 ( n ) \displaystyle \text{log}_{n}(3) = \frac{1}{\text{log}_{3}(n)} which tends to 0 0 as n → ∞ n\to\infty . So let us find N ( ϵ ) N(\epsilon) . we have ∣ log n ( 3 ) ∣ < ϵ ⟹ log n ( 3 ) < ϵ ⟹ n > 3 1 ϵ \displaystyle \lvert\text{log}_{n}(3)\rvert <\epsilon \implies \text{log}_{n}(3)<\epsilon \implies n> 3^{\frac{1}{\epsilon}} . Hence N ( ϵ ) N(\epsilon) works for A A .

For the 2nd limit we see that the limit is lim ⁡ n → ∞ n n ⋅ 1 1 + 1 n 2 = 1 \displaystyle \lim_{n\to\infty}\frac{n}{n}\cdot\frac{1}{\sqrt{1+\frac{1}{n^{2}}}}=1 . Also we have a n < 1 ∀ n ∈ N a_{n} <1\,\,\,\,\,\forall\,n\in\N So we have ∣ a n − 1 ∣ = 1 − a n = 1 − n n 2 + 1 = n 2 + 1 − n n 2 + 1 = ( n 2 + 1 − n ) ( n 2 + 1 + n ) ( n 2 + 1 ) ( n 2 + 1 + n ) = 1 ( n 2 + 1 ) ( n 2 + 1 + n ) < 1 2 n 2 \displaystyle \lvert a_{n}-1\rvert = 1-a_{n} = 1-\frac{n}{\sqrt{n^{2}+1}} = \frac{\sqrt{n^{2}+1} -n}{\sqrt{n^{2}+1}} =\frac{(\sqrt{n^{2}+1} -n)(\sqrt{n^{2}+1} +n)}{(\sqrt{n^{2}+1})(\sqrt{n^{2}+1} +n)}=\frac{1}{(\sqrt{n^{2}+1})(\sqrt{n^{2}+1}+n)} <\frac{1}{2n^{2}}

Hence ϵ > 1 2 n 2 ⟹ n > 1 2 ϵ \epsilon>\frac{1}{2n^{2}} \implies n>\frac{1}{\sqrt{2\epsilon}} . So N ( ϵ ) N(\epsilon) works in this case also.Hence B B is a valid pair.

Now let us check for the last case :-

lim ⁡ n → ∞ a n = lim ⁡ n → ∞ 2 n n 2 + 2 n + n = 1. \displaystyle \lim_{n\to\infty} a_{n} = \lim_{n\to\infty}\frac{2n}{\sqrt{n^{2}+2n}+n} = 1.

Also it is always less than 1.

So let us examine 1 − a n = n 2 + n − n n 2 + n + n = 2 n 2 n 2 + 2 n + 2 n n 2 + n = n n 2 + n + n n 2 + 1 \displaystyle 1-a_{n} = \frac{\sqrt{n^{2}+n}-n}{\sqrt{n^{2}+n}+n} = \frac{2n}{2n^{2}+2n+2n\sqrt{n^{2}+n}}=\frac{n}{n^{2}+n+n\sqrt{n^{2}+1}} which is O ( 1 n ) O(\frac{1}{n}) where O ( . ) O(.) denotes the big O notation .

But let us examine the function N ( ϵ ) N(\epsilon) given .

we have n > ( 1 − ϵ ) 2 10 ϵ 2 ⟹ ϵ > 1 10 n + 1 n>\frac{(1-\epsilon)^{2}}{10\epsilon^{2}} \implies \epsilon>\frac{1}{\sqrt{10n}+1} which is O ( 1 n ) O(\frac{1}{\sqrt{n}}) .

Hence n > N ( ϵ ) n>N(\epsilon) is not enough to ensure 1 − a n < ϵ 1-a_{n} < \epsilon . Hence C C does not work.

Here the logic is simple.... For a given ϵ \epsilon and corresponding to it the natural number N ( ϵ ) N(\epsilon) as defined in the question , n > N ( ϵ ) n>N(\epsilon) would not ensure that ϵ > n n 2 + n + n n 2 + 1 \epsilon >\frac{n}{n^{2}+n+n\sqrt{n^{2}+1}} because the function n n 2 + n + n n 2 + 1 \frac{n}{n^{2}+n+n\sqrt{n^{2}+1}} decreases to 0 0 more rapidly than 1 10 n + 1 \frac{1}{\sqrt{10n}+1} as n goes to infinity.

2 Helpful 1 Interesting 2 Brilliant 0 Confused

Thank you for the solution provided. I solved the first limit in the same way, the third in a similar and the second in a different way and got only A correct. Can you spot the mistake in my solution or how does it differ from your one? lim ⁡ n → ∞ a n = 1 \lim_{n\to\infty}a_n=1 That's easy. Then we spot that n n 2 + 1 \frac{n}{\sqrt{n^2+1}} is always less than 1 and ∣ a n − 1 ∣ = 1 − a n |a_n-1|=1-a_n . And then simply solve the inequality: 1 − n n 2 + 1 < ϵ 1-\frac{n}{\sqrt{n^2+1}}<\epsilon 1 − ϵ < n n 2 + 1 1-\epsilon<\frac{n}{\sqrt{n^2+1}} 1 − ϵ < 1 1 + 1 n 2 1-\epsilon<\frac{1}{\sqrt{1+\frac{1}{n^2}}} 1 1 − ϵ > 1 + 1 n 2 \frac{1}{1-\epsilon}>\sqrt{1+\frac{1}{n^2}} 1 ( 1 − ϵ ) 2 > 1 + 1 n 2 \frac{1}{(1-\epsilon)^2}>1+\frac{1}{n^2} 1 n 2 < 1 ( 1 − ϵ ) 2 − 1 = 1 − 1 + 2 ϵ − ϵ 2 ( 1 − ϵ ) 2 = 2 ϵ − ϵ 2 ( 1 − ϵ ) 2 \frac{1}{n^2}<\frac{1}{(1-\epsilon)^2}-1=\frac{1-1+2\epsilon-\epsilon^2}{(1-\epsilon)^2}=\frac{2\epsilon-\epsilon^2}{(1-\epsilon)^2} n 2 > ( 1 − ϵ ) 2 2 ϵ − ϵ 2 n^2>\frac{(1-\epsilon)^2}{2\epsilon-\epsilon^2} n > 1 − ϵ 2 ϵ − ϵ 2 = N ( ϵ ) ≠ 1 2 ϵ n>\frac{1-\epsilon}{\sqrt{2\epsilon-\epsilon^2}}=N(\epsilon)\neq \frac{1}{\sqrt{2\epsilon}} That's why I picked the answer A only.

Veselin Dimov - 5 months, 4 weeks ago

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@Veselin Dimov .The N ( ϵ ) N(\epsilon) is just a function of epsilon which satisfies the inequality,it does not mean that the function has to be unique . One can find many such N ( ϵ ) N(\epsilon) which would satisfy the inequality. The question does not want you to find such an N ( ϵ ) N(\epsilon) but rather it asks you to check whether the pairs of a n a_{n} and N ( ϵ ) N(\epsilon) as defined by the op is suitable. In the 2nd case the function you have derived for N works as well as the one defined by the user.

For example....what I have shown is that ∣ a n − 1 ∣ < ϵ \lvert a_{n} - 1 \rvert <\epsilon when n > 1 2 ϵ n>\frac{1}{\sqrt{2\epsilon}} . but let us go a step back . we have 1 ( n 2 + 1 ) ( n 2 + 1 + n ) < 1 2 n 2 \displaystyle \frac{1}{(\sqrt{n^{2}+1})(\sqrt{n^{2}+1}+n)} <\frac{1}{2n^{2}} . But we can go even further .

1 ( n 2 + 1 ) ( n 2 + 1 + n ) < 1 2 n 2 < 1 2 n \displaystyle \frac{1}{(\sqrt{n^{2}+1})(\sqrt{n^{2}+1}+n)} <\frac{1}{2n^{2}}<\frac{1}{2n} which means what N ( ϵ ) = 1 2 ϵ N(\epsilon) = \frac{1}{2\epsilon} would work perfectly fine.

Or even going a step further:-

1 ( n 2 + 1 ) ( n 2 + 1 + n ) < 1 2 n 2 < 1 2 n < 1 2 n \displaystyle \frac{1}{(\sqrt{n^{2}+1})(\sqrt{n^{2}+1}+n)} <\frac{1}{2n^{2}}<\frac{1}{2n}<\frac{1}{2\sqrt{n}} would give us N ( ϵ ) = 1 4 ϵ 2 \displaystyle N(\epsilon)=\frac{1}{4\epsilon^{2}} . So this works as well. In this way you can create an infinite number of functions N ( ϵ ) N(\epsilon) which would satisfy the inequality.

What you have done is also find one such N ( ϵ ) N(\epsilon) which works , but the question simply wants you to check whether the N ( ϵ ) N(\epsilon) defined by the problem maker works or not. The point here to note is that since ∣ a n − 1 ∣ \lvert a_{n} - 1 \rvert is O ( 1 n 2 ) O(\frac{1}{n^{2}}) ,so what we cannot do is take a function of n which decreases faster than ∣ a n − 1 ∣ \lvert a_{n} - 1 \rvert . So for example we cannot take 1 ϵ 3 \displaystyle \frac{1}{\sqrt[3]{\epsilon}} because it gives ϵ > 1 n 3 \displaystyle \epsilon > \frac{1}{n^{3}} which may not hold for some n . This was my reasoning for the third option.

Arghyadeep Chatterjee - 5 months, 4 weeks ago

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Thank you for the comprehensive answer, very nice explained

Veselin Dimov - 5 months, 3 weeks ago

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