# Thread: Calculus & Analysis Marathon & Questions

1. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
Yeah basically, but part a) could've been smacked straight away with the first FTC and the chain rule

Hiccup on the final answer for d) though
wait what?
I don't know if it's legit lol

2. ## Re: First Year Uni Calculus Marathon

Originally Posted by integral95
wait what?
I don't know if it's legit lol
f'(1) = 3√17 not 3√5

3. ## Re: First Year Uni Calculus Marathon

$\noindent Let P and Q be non-constant polynomials with positive leading coefficients. Show that \lim _{y\to \infty}\frac{P\left(y\right)}{Q\left(\ln y\right)}=\infty.$

4. ## Re: First Year Uni Calculus Marathon

I'm taking a massively blank stab at this one for my procrastination from accounting, so the quality of this answer will be again, poor lol.

$If \deg{P}\ge \deg{Q}, then \\L=\lim_{y \to \infty}{\frac{a_{n+a}y^{n+a}+a_{n+a-1}y^{n+a-1}+\dots+a_ny^n+a_{n-1}y^{n-1}+\dots+a_0}{b_n{(\ln{y}) }^n + b_{n-1}(\ln{y}) ^{n-1} + \dots + b_0 } }$

$Then, just the fact that as x tends to \infty, \\ \ln{x}=o(x) and consequently (\ln{x})^n=o\left(x^n\right) can be used to show that the numerator increases faster than the denominator, i.e. the limit diverges.\\ Note: Obviously y^{n+a}>y^n for large y if n \ge 1$

$If \deg{P} < \deg{Q} however, note that y^m \to \infty, (\ln{y})^n \to \infty as y \to \infty so we can use L'Hopital's rule (for any m>0, n>0)$

\begin{align*} \lim_{y \to \infty}{\frac{P(y)}{Q(\ln{y})}} &\stackrel{L'H}{=} \lim_{y \to \infty}{\frac{P^\prime (y)}{Q^\prime (\ln{y}) \cdot \frac{1}{y} } } \\ &= \lim_{y \to \infty}{\frac{y P^\prime (y)}{Q^\prime (\ln{y}) } } \\ &\stackrel{L'H}{=} \lim_{y \to \infty}{\frac{y P^{\prime \prime} (y) + P^\prime (y)}{Q^{\prime \prime} (\ln{y}) \cdot \frac{1}{y} } } \\ &= \lim_{y \to \infty}{\frac{y^2 P^{\prime \prime} (y) + y P^\prime (y)}{Q^{\prime \prime} (\ln{y}) } } \\ &\stackrel{L'H}{=} \lim_{y \to \infty}{\frac{y^3 P^{\prime \prime \prime} (y) +3y^2 P^{\prime \prime} (y) + y P^\prime (y)}{Q^{\prime \prime \prime} (\ln{y}) } } \end{align*}\\ \text{and so on}

$Interestingly, with each application of L'Hopital's rule, the degree of the numerator stays the same, but the degree of the denominator decreases by 1. This is because the \frac{1}{y} introduced into the denominator as per each application of L'H raises the degree of the numerator back up by one, each time that decreases (hence, overall effects cancel out). After a certain amount of derivatives being taken, the denominator will become a constant term, which will obviously imply the limit tends to infinity.$

5. ## Re: First Year Uni Calculus Marathon

Wouldn't you have to also check/assume the individual requirements of L'Hs are satisfied, namely the limits have to all exist?

6. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
$\noindent Let P and Q be non-constant polynomials with positive leading coefficients. Show that \lim _{y\to \infty}\frac{P\left(y\right)}{Q\left(\ln y\right)}=\infty.$
Change variables.

P(ex)/Q(x)

Without Loss of Generality, we can assume P is of degree 1.

This expression obviously tends to infinity.

For those who do not see it, apply L'Hôpital's Rule (with respect to x) n times where n is the degree of Q.

If P is larger, then the contribution to the explosion is greater, and the expression still diverges.

[QED]

7. ## Re: First Year Uni Calculus Marathon

Change variables.

P(ex)/Q(x)

Without Loss of Generality, we can assume P is of degree 1.

This expression obviously tends to infinity.

For those who do not see it, apply L'Hôpital's Rule (with respect to x) n times where n is the degree of Q.

If P is larger, then the contribution to the explosion is greater, and the expression still diverges.

[QED]
How are you able to do these uni questions when you're not even at uni?

8. ## Re: First Year Uni Calculus Marathon

Originally Posted by BlueGas
How are you able to do these uni questions when you're not even at uni?
These things are fairly simple to understand, you don't have to be taking a high level mathematics course to be able to do them. I'm sure I could find people in year 10 who can do these questions.

9. ## Re: First Year Uni Calculus Marathon

Originally Posted by dan964
Wouldn't you have to also check/assume the individual requirements of L'Hs are satisfied, namely the limits have to all exist?
That's a statement I forgot to (explicitly) make.

But I obviously assumed so because the final limit is supposed to exist and tend to infinity

10. ## Re: First Year Uni Calculus Marathon

These things are fairly simple to understand, you don't have to be taking a high level mathematics course to be able to do them. I'm sure I could find people in year 10 who can do these questions.
That's abit of an exaggeration isn't it? Would you find people in Year 10 (unless if they accelerate) that understand L'Hôpital's Rule?

11. ## Re: First Year Uni Calculus Marathon

Originally Posted by BlueGas
That's abit of an exaggeration isn't it? Would you find people in Year 10 (unless if they accelerate) that understand L'Hôpital's Rule?
Yes. I would, and I can, and I will. I am quasi-mentoring (mainly because he is very self capable) a year 10 student who has self-taught all the basics of Extension 2 Integration, and I will be teaching him competitive integration. Then I will drag him to the MATHSOC Integration Bee next year.

12. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
That's a statement I forgot to (explicitly) make.

But I obviously assumed so because the final limit is supposed to exist and tend to infinity
Limits that tend to infinity, do they actually exist in the reals?

13. ## Re: First Year Uni Calculus Marathon

Going to infinity counts as "exist" for the purposes of using L'Hôpital's rule. You can't have something that oscillates forever though (like (2+cos(x))/(2-cos(x)) or something), that'd fail to satisfy the requirement for L'Hôpital's rule.

14. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
Going to infinity counts as "exist" for the purposes of using L'Hôpital's rule. You can't have something that oscillates forever though (like (2+cos(x))/(2-cos(x)) or something), that'd fail to satisfy the requirement for L'Hôpital's rule.
Identically equal to making a manipulation/substitution so that you get 0/0 instead of ∞/∞

15. ## Re: First Year Uni Calculus Marathon

$f(x) = \begin{cases} \tan^{-1}{x} \quad \text{if } x \text{ is rational}\\ \sin{x} \quad \quad \, \text{if }x\text{ is irrational} \end{cases}$

Prove that f is differentiable at the origin.

There is a real positive number θ which satisfies tan-1θ = sinθ.

Prove that f is continuous at θ.

Hence prove f is continuous at -θ.

16. ## Re: First Year Uni Calculus Marathon

Yes. I would, and I can, and I will. I am quasi-mentoring (mainly because he is very self capable) a year 10 student who has self-taught all the basics of Extension 2 Integration, and I will be teaching him competitive integration. Then I will drag him to the MATHSOC Integration Bee next year.
...just saw (realised) this

17. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
...just saw (realised) this
cool

18. ## Re: First Year Uni Calculus Marathon

$\noindent Let f\colon\mathbb{R}\to \mathbb{R} be a function that is continuous on an open interval I containing a\in \mathbb{R} and is differentiable on I, except possible at a. Suppose \lim_{x\to a}f^{\prime}\left(x\right) exists. Does f^\prime \left(a\right) necessarily exist? Justify your answer.$

19. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
$\noindent Let f\colon\mathbb{R}\to \mathbb{R} be a function that is continuous on an open interval I containing a\in \mathbb{R} and is differentiable on I, except possible at a. Suppose \lim_{x\to a}f^{\prime}\left(x\right) exists. Does f^\prime \left(a\right) necessarily exist? Justify your answer.$
Yep. MVT implies (f(x)-f(a))/(x-a)=f'(c(x)) for some c(x) strictly between a and x. As x->a, c(x)->a, which implies that the RHS converges to a limit by the question's assumptions. So the LHS converges to a limit as x->a, which is precisely the definition of differentiability.

A related followup:
Let f:R->R be a differentiable function. Must f' be continuous?

20. ## Re: First Year Uni Calculus Marathon

Originally Posted by seanieg89
Yep. MVT implies (f(x)-f(a))/(x-a)=f'(c(x)) for some c(x) strictly between a and x. As x->a, c(x)->a, which implies that the RHS converges to a limit by the question's assumptions. So the LHS converges to a limit as x->a, which is precisely the definition of differentiability.

A related followup:
Let f:R->R be a differentiable function. Must f' be continuous?
Going by memory isn't the function f(x)=x. sin(1/x) for x \neq 0 ; 0 for x=0 a case where f is differentiable everywhere but f' is not continuous at 0?

21. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
Going by memory isn't the function f(x)=x. sin(1/x) for x \neq 0 ; 0 for x=0 a case where f is differentiable everywhere but f' is not continuous at 0?
You should never go just by memory!

Do you believe that this is a counterexample? And if so, why?

22. ## Re: First Year Uni Calculus Marathon

And another related Q.

$\noindent Does there exist a continuous function f \colon \mathbb{R}\to \mathbb{R} with \lim_{x \to a}f^\prime \left(x\right) existing (for some real number a) but not equal to f^\prime (a)? (seanieg89 has just shown above that f^\prime (a) exists.) Justify your answer.$

23. ## Re: First Year Uni Calculus Marathon

Originally Posted by seanieg89
You should never go just by memory!

Do you believe that this is a counterexample? And if so, why?
Ouch! Maybe I meant x2 sin(1/x). Anyway here's an elementary proof

$\text{To verify }f\text{ is continuous at }0\text{, show that }\lim_{x\to 0}x^2 \sin{\frac{1}{x}}=0\text{ using the squeeze theorem on the range of sine.}\\ \text{Or basically taking limits to 0 on }-x^2 \le x^2\sin{\frac{1}{x}} \le x^2$

$\text{Similarly it can be shown by the definition of the derivative that }f^\prime (0)=\lim_{ h \to 0}h \sin \frac{1}{h}=0 \text{ using the same thing.}$

I think because f follows the same rule from the left and the right there is no need to take 1-sided limits here.

$\text{So we have} \\ f^\prime(x)=\begin{cases}2x \sin \frac{1}{x}-2\cos \frac{1}{x} &, x\neq 0 \\ 0 & ,x=0\end{cases}$

$\text{But }\lim_{x \to 0^+}f^\prime (x)\text{ does not exist as you can't find }\lim_{x \to 0^+}\cos{\frac{1}{x}}$

$\text{Rather, }f^\prime\text{ has an essential discontinuity at }0$

24. ## Re: First Year Uni Calculus Marathon

Bingo. Haha I figured you would realise that the previous exponent wasn't good enough when you tried to work out the details.

25. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
And another related Q.

$\noindent Does there exist a continuous function f \colon \mathbb{R}\to \mathbb{R} with \lim_{x \to a}f^\prime \left(x\right) existing (for some real number a) but not equal to f^\prime (a)? (seanieg89 has just shown above that f^\prime (a) exists.) Justify your answer.$
Doesn't my proof of existence show that f'(a)=L, the assumed limit of f'(x) as x->a? We showed the function was differentiable at a by finding it's derivative at a, which was L.

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