a. Assuming that limn →∞(1 / n^c)=0 if c is any positive constant, show that limn →∞ (lnn)/(n^c)

step 1 So we have two things in this question we want to do. So in part A, we want to show that the lim

A. Assuming that limn →∞(1 / n^c)=0 if c is any positive...

a. Assuming that $\lim _{n \rightarrow \infty}\left(1 / n^{c}\right)=0$ if $c$ is any positive constant, show that $$ \lim _{n \rightarrow \infty} \frac{\ln n}{n^{c}}=0 $$ if $c$ is any positive constant. b. Prove that $\lim _{n \rightarrow \infty}\left(1 / n^{\prime}\right)=0$ if $c$ is any positive constant. (Hint: If $\varepsilon=0.001$ and $c=0.04,$ how large should $N$ be to ensure that $\left|1 / n^{c}-0\right|<\varepsilon$ if $n>N ? )$

a. Assuming that $\lim _{n \rightarrow \infty}\left(1 / n^{c}\right)=0$ if $c$ is any positive constant, show that
$$
\lim _{n \rightarrow \infty} \frac{\ln n}{n^{c}}=0
$$
if $c$ is any positive constant.
b. Prove that $\lim _{n \rightarrow \infty}\left(1 / n^{\prime}\right)=0$ if $c$ is any positive constant.
(Hint: If $\varepsilon=0.001$ and $c=0.04,$ how large should $N$ be to
ensure that $\left|1 / n^{c}-0\right|<\varepsilon$ if $n>N ? )$

Key Concepts

Limits and Convergence

This concept involves understanding the behavior of functions or sequences as their input grows without bound. In the context of real analysis, proving that a sequence converges to a particular number (in this case, zero) means demonstrating that for every positive epsilon, there is a point beyond which all sequence terms fall within epsilon of the target value.

Asymptotic Growth Rates

Asymptotic growth rates refer to comparing how fast different types of functions increase as their variable approaches infinity. In many situations, functions involving powers (or polynomial growth) will eventually dominate logarithmic functions, no matter how small the exponent is. This fact is critical when proving that the ratio of a logarithm to any positive power of n tends to zero.

Epsilon-N Definition of Convergence

This formal definition is central to proofs in analysis. It states that a sequence converges to a limit L if, for every epsilon greater than zero, there exists a natural number N such that for all n greater than N, the absolute difference between the sequence’s term and L is less than epsilon. This method is key in demonstrating the convergence of the provided sequences.

Comparison of Function Growth Rates

This key concept involves evaluating which functions increase more rapidly as their variable grows larger. Understanding that logarithmic functions grow slower than any positive power function allows one to show that when the logarithm is in the numerator and a power function in the denominator, the resulting ratio eventually becomes arbitrarily small, leading to a limit of zero.

Transcript

00:01 So we have two things in this question we want to do.

00:03 So in part a, we want to show that the limit as in approaches infinity of the natural log of n over in to the c is equal to zero when c is some positive constant.

00:14 And we want to do this using that the limit as in approaching infinity of 1 over nc is equal to 0.

00:21 So let's go ahead and start by just writing out the limit as an approach is infinity of the natural log of 0.

00:31 Over into the c.

00:34 We'll notice if we were to just evaluate this directly, we know the natural log of n goes to infinity, and we know that n goes to infinity, and if we raise that to some positive constant, it will still go to infinity.

00:50 So this tells us we can apply lopetal's rule, since we have infinity over infinity, so lopatals, and so now we can go ahead and take the derivative of what we have in the numerator so the derivative of natural log is going to be 1 over n and then we take the derivative of n c which we would need to use the power rule to do so we'd have c n minus so in into the c minus 1 and now i'm going to pull this c constant out front so it's 1 over c times the limit as an approach of oh well one over n over n divided by into the c, well that would just move that into the denominator and we could rewrite that as into the c.

01:46 And we are just assuming that the limit as an approaches infinity of 1 over into the c is 0.

01:54 So we can go ahead and plug this in and using that we get 1 over c times 0 and then 1 times c or 1 over c times 0 is 0.

02:10 And so we've proved that this here does actually approach zero.

02:20 Now for part b, we want to show that this limit that we assume to be true, well, we want to actually show that.

02:30 So let's go ahead and do a little bit of scratch work first.

02:35 So scratch.

02:38 So remember, what we want to show is that for every upslown greater than zero, so there exists some capital n such that when little n is strictly bigger than capital n, a .n minus l absolute value will be less than that epsilon.

02:58 So the thing we really want to show is that we can find this capital n here.

03:04 So the way we would normally go about this is start with what we want to be true and then work backwards.

03:11 So, a, n, in this case, is going to be 1 over n to z.

03:16 Our limit, we're assuming, is 0, and we want this to just be less than epsilon.

03:23 Now we can go ahead and drop the absolute value, since 1 over into the c minus 0, would just be 1 over into the c...

Please give Ace some feedback