Proving limits for roots and exponents

[Lambda96]

Homework Statement

See post

Relevant Equations

none

Hi

I have to prove the following three tasks

I now wanted to prove three tasks with a direct proof, e.g. for task a)$$\sqrt[n]{n} = n^{\frac{1}{n}}= e^{ln(n^{\frac{1}{n}})}=e^{\frac{1}{n}ln(n)}$$
$$\displaystyle{\lim_{n \to \infty}} \sqrt[n]{n}= \displaystyle{\lim_{n \to \infty}} e^{\frac{1}{n}ln(n)}$$

I would now argue as follows that x tends to infinity faster than the logarithm and therefore $\frac{1}{n}$ tends to zero and therefore $e^0=1$.

Would this be a valid proof for task a?

 

[Lambda96]

Unfortunately my calculations are not displayed via Latex, does anyone know what the reason for this is? I have tried $$$ $ and ### # but with both the calculations are not displayed.

With overleaf the calculations are displayed

 

[berkeman]

Double-$ is the correct delimiter (before and after) for stand-alone lines of Latex, and double-# is correct for in-line LaTeX.

So far I haven't been able to figure out what is wrong with your LaTeX, so I've reported it to the other Mentors. In the mean time, have you looked through the "LaTeX Guide" link below the Edit window to see if you see any issues?

 

[Hill]

Here you go:$$\sqrt[n]{n}= n^{\frac{1}{n}}=e^{ln(n^{\frac{1}{n}})}=e^{\frac{1}{n}ln(n)}$$ $$\lim_{n \rightarrow \infty} \sqrt[n]{n}= \lim_{n \rightarrow \infty} e^{\frac{1}{n}ln(n)}$$

 

[Mark44]

Unfortunately my calculations are not displayed via Latex

The LaTeX is now fixed. You had some obscure LaTeX commands in your expressions that probably aren't part of the MathJax that we use here; namely uproot and leftroot. In addition the lines that wouldn't render correctly were missing a right brace ( } ) at the end.
In any case uproot and leftroot weren't needed for nth roots. Use $\sqrt[n] {n}$ (raw ##\sqrt[n] {n}##) instead.

 

[fresh_42]

Homework Statement: See post
Relevant Equations: none

Hi

I have to prove the following three tasks

View attachment 336035
I now wanted to prove three tasks with a direct proof, e.g. for task a)$$\sqrt[n]{n} = n^{\frac{1}{n}}= e^{ln(n^{\frac{1}{n}})}=e^{\frac{1}{n}ln(n)}$$
$$\displaystyle{\lim_{n \to \infty}} \sqrt[n]{n}= \displaystyle{\lim_{n \to \infty}} e^{\frac{1}{n}ln(n)}$$

I would now argue as follows that x tends to infinity faster than the logarithm and therefore $\frac{1}{n}$ tends to zero and therefore $e^0=1$.

Would this be a valid proof for task a?

Basically yes, but I would require an argument for $\lim_{n \to \infty}\dfrac{\log n}{n}=0$ likely with the help of L'Hôpital.

 

[WWGD]

If you've seen this:
For f continuous,
$Lim _{x\rightarrow a} f(x)=f(a)$
You can use it too.
Edit: It would most likely require L'Hopital too.

 

[Lambda96]

Thank you berkeman, Hill, Mark44, fresh_42 and WWGD for your help

I then used de L'Hospital's rule to show that the term $\lim_{n \rightarrow \infty} \frac{\ln(n)}{n}$ goes to zero so

$$\lim_{n \rightarrow \infty} \frac{\ln(n)}{n}=\lim_{n \rightarrow \infty} \frac{1}{n}=0$$Fortunately, I didn't have any problems with task b, but I can't make any progress at all with task c.

My lecturer told us the following for the solution$$\lim_{n \rightarrow \infty} n^s = \infty \quad \text{for} \, s \in \mathbb{Q}_{+}$$

$$\lim_{n \rightarrow \infty} a^n = \infty \quad \text{for} \, a > 1$$

Unfortunately, I can't do anything with this because I already knew that the two terms tend to infinity. Would I have to show somehow that $a^n$ tends to infinity faster than $n^s$?

 

[PeroK]

My lecturer told us the following for the solution$$\lim_{n \rightarrow \infty} n^s = \infty \quad \text{for} \, s \in \mathbb{Q}_{+}$$

$$\lim_{n \rightarrow \infty} a^n = \infty \quad \text{for} \, a > 1$$

Unfortunately, I can't do anything with this because I already knew that the two terms tend to infinity. Would I have to show somehow that $a^n$ tends to infinity faster than $n^s$?

There are a lot of possible cases in part c). Perhaps try $s \in \mathbb N$ first?

 

[PeroK]

Hint to simplify things:$$\lim_{n \rightarrow \infty} c_n = 0 \ \Leftrightarrow \ \lim_{n \rightarrow \infty} |c_n| = 0$$

 

[fresh_42]

Fortunately, I didn't have any problems with task b, but I can't make any progress at all with task c.

I haven't done c) so my advice must be taken as an idea. not as a solution path.

I would start (as usual) with what we already know. We have $s=\dfrac{p}{q}\in \mathbb{Q}$ and $|a|>1$ that leads to two cases $a=1+r$ and $a=-1-r$ with $r>0.$

The first case is thus $\lim_{n \to \infty}\dfrac{\sqrt[q]{n^p}}{(1+r)^n}$ and I would look up the Taylor series for $\sqrt[q]{\cdot}$ at infinity on WA. I'm starting to see what @PeroK meant with a lot of cases ...

Edit: With the hint in post #10, we only have to show the case $a>1.$ This means we need to show that $$\left|\dfrac{\sqrt[q]{n^p}}{(1+r)^n}\right|<\varepsilon $$

 

[pasmith]

I don't know what the purpose of the restriction to $s \in \mathbb{Q}$ in part (c) is, since I think the result is the same for $s \in \mathbb{R} \setminus \mathbb{Q}$. Possibly the restriction is there because $x^\alpha$ for irrational $\alpha$ and positive $x$ has not yet been defined (as $\exp(\alpha \ln x)$). If so, the (to me) obvious $$\lim_{n \to \infty} \left|\frac{n^s}{a^n}\right| = \lim_{n \to \infty} \exp( s \ln n - n \ln |a|)$$ may not be admissible. But you've already used $n^{1/n} = \exp(n^{-1}\ln n)$ in part (a), so...

 

[WWGD]

I don't know what the purpose of the restriction to $s \in \mathbb{Q}$ in part (c) is, since I think the result is the same for $s \in \mathbb{R} \setminus \mathbb{Q}$. Possibly the restriction is there because $x^\alpha$ for irrational $\alpha$ and positive $x$ has not yet been defined (as $\exp(\alpha \ln x)$). If so, the (to me) obvious $$\lim_{n \to \infty} \left|\frac{n^s}{a^n}\right| = \lim_{n \to \infty} \exp( s \ln n - n \ln |a|)$$ may not be admissible. But you've already used $n^{1/n} = \exp(n^{-1}\ln n)$ in part (a), so...

I think it has to see with defining a (edit: Continuous)function on a dense subset ( of the Reals), which defines it uniquely.

 

[Lambda96]

Thank you PeroK, fresh_42, pasmith and WWGD for your help

Since many other students had problems with the exercise, our lecturer told us to use the epsilon criterion for the proof.

I then used fresh_42's suggestion and estimated the result, hope it was correct, the submission was today