# Thread: Calculus & Analysis Marathon & Questions

1. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
Believe it or not my workbook was happy if you just used L'Hopitals
Not rigorous enough :P

jks. But I cringe at directly using that on the expression, I would rather use it on the bounds, it's much more nice-looking.

2. ## Re: First Year Uni Calculus Marathon

@IG

Still suspecting I haven't fully finished seanieg's question though. Anything else to pick up?

3. ## Re: First Year Uni Calculus Marathon

^ I think you should try invoking the IVT to prove some things I think you're trying to prove.

4. ## Re: First Year Uni Calculus Marathon

(Sorry about delayed response, am wrestling with a problem of my own )

Yep @leehuan, IVT, and EVT are key to rigorous argument here. You are on a nice train of thought and I think you have a good intuitive grasp of why k=2 is impossible, but making this rigorous takes some care.

The "infinitesimal" part of your proof in particular needs to be made more precise, and things like Rolle's theorem are inapplicable here because the function need not be differentiable.

From what you are saying though, it seems to me you have the correct "picture" in your head for your k=2 disproof.

(I am also unconvinced that this "picture" serves to disprove the k>2 possibility so think about explaining that a little clearer).

5. ## Re: First Year Uni Calculus Marathon

But what about k = ∞ ?
Yes it is possible. Eg: f(x)=x*sin(x).

Originally Posted by porcupinetree
Part 1 (not a very rigorous proof, though):

$As x\rightarrow \infty , x^3 + 1 \rightarrow x^3$
$As x\rightarrow \infty , x^2 + 1 \rightarrow x^2$
$Thus, \lim_{x \rightarrow \infty} \frac{\ln (x^3 + 1)}{\ln (x^2 + 1)} = \frac{\ln(x^3)}{\ln(x^2)} = \frac{3\ln(x)}{2\ln(x)} = \frac{3}{2}$
This is easy to make rigorous. For large x, x^3+1 < 2x^3 and x^2+1 < 2x^2. So

log(x^3+1)/log(x^2+1) < log(2x^3)/log(x^2) = (log(2)+3log(x))/2log(x).

and

log(x^3+1)/log(x^2+1) > log(x^3)/log(2x^2)=3log(x)/(log(2)+2log(x)).

Squeeze to finish.

Originally Posted by leehuan
Hmm...

$Cases where differentiability arise out of a discontinuity can be safely disregarded as the question already specified f is continuous on \mathbb R$

$This means that the lack of differentiability arises in the event of a cusp or a corner. However, that same cusp/corner will essentially also be a local extrema, thus the proof proceeds from there in a similar manner.$

At least, I don't think a cusp/corner can't be a local extrema here?
Corners do not have to be local extrema. Eg f(x)=max(x,2x) at 0.

Also, lack of differentiability can occur in ways worse than corners. (f(x)-f(a))/(x-a) needs not tend to a limit at all as x tends to a, or it might only tend to a limit from one side.

It is more trouble than it is worth to appeal to theorems about differentiable functions and then treat the exceptional points separately. (In fact a continuous function might be differentiable nowhere!)

6. ## Re: First Year Uni Calculus Marathon

Hmmm

With the IVT I can pretty much find how to apply it. Just get rid of that infinitesimal stuff and properly redefine it (which I might do later). Which gives me some ideas on the EVT but I haven't placed too much thought into it to know what's right and what's flawed.

May continue this when I have more free time. Gah I hate assignment based subjects

7. ## Re: First Year Uni Calculus Marathon

Originally Posted by seanieg89
Yes it is possible. Eg: f(x)=x*sin(x).

This is easy to make rigorous. For large x, x^3+1 < 2x^3 and x^2+1 < 2x^2. So

log(x^3+1)/log(x^2+1) < log(2x^3)/log(x^2) = (log(2)+3log(x))/2log(x).

and

log(x^3+1)/log(x^2+1) > log(x^3)/log(2x^2)=3log(x)/(log(2)+2log(x)).

Squeeze to finish.

Corners do not have to be local extrema. Eg f(x)=max(x,2x) at 0.

Also, lack of differentiability can occur in ways worse than corners. (f(x)-f(a))/(x-a) needs not tend to a limit at all as x tends to a, or it might only tend to a limit from one side.

It is more trouble than it is worth to appeal to theorems about differentiable functions and then treat the exceptional points separately. (In fact a continuous function might be differentiable nowhere!)
*cough*weierstrass*cough*

8. ## Re: First Year Uni Calculus Marathon

*cough*weierstrass*cough*
Imagine Carrotsticks BOS trial last Q. of 4U paper 2016: 'Sketch the graphs of the Weierstrass and Dirichlet functions' .

9. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
Imagine Carrotsticks BOS trial last Q. of 4U paper 2016: 'Sketch the graphs of the Weierstrass and Dirichlet functions' .
last part of the question

"also note significant points on the graph"

10. ## Re: First Year Uni Calculus Marathon

Originally Posted by InteGrand
Imagine Carrotsticks BOS trial last Q. of 4U paper 2016: 'Sketch the graphs of the Weierstrass and Dirichlet functions' .
the final part will be:

hence, sketch the indefinite integral of the weierstrass and dirichlet functions, with any choice of C you like.

11. ## Re: First Year Uni Calculus Marathon

Originally Posted by porcupinetree
$Next question:$
$Show that f, where f is defined by$
$f(x) = \left\{\begin{matrix} x^3 & if x is rational \\ 0 & if x is irrational \end{matrix}\right.$
$is differentiable at x=0.$
Regarding the continuity part:

Informally, when x≥0, 0≤f(x)≤x3 and when x≤0, x3≤f(x)≤0

But in both cases if we take limits to nought, by the squeeze theorem we have lim x->0 f(x) = 0 which incidentally enough equals f(0)
______________________

The answer given to the derivative was wrong because I didn't regard the significance of what h was.

lim x->0 [ f(0+h)-f(0) ] / h
= lim x->0 f(h)/h

but f(h) = h3 when h is rational
and f(h) = 0 when h is irrational

Note then that f(h)/h = h2 for h in Q
or = 0 for h not in Q

Haven't done questions like this one before yet but just staring at it I'm inclined to say the squeeze theorem can be used again

12. ## Re: First Year Uni Calculus Marathon

Yeah it's basically squeeze law. We can use that to show that f'(0) = 0.

13. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
Haven't done questions like this one before yet but just staring at it I'm inclined to say the squeeze theorem can be used again
$\left|\frac{f(x)-f(0)}{x-0}\right|=\frac{|f(x)|}{|x|}\leq \frac{|x|^3}{|x|}=|x|^2\rightarrow 0\\ \\ \Rightarrow \frac{f(x)-f(0)}{x-0}\rightarrow 0.$

(The "=>" comes from the squeeze law. Generally if we suspect something might tend to zero, it is often convenient to take absolute values as then we just need to bound it above by something that tends to zero.)

14. ## Re: First Year Uni Calculus Marathon

$\lim_{x \rightarrow 0} \left(\frac{1+\tan{x}}{1+\sin{x}}\right)^{\frac{1} {\sin{x}}}$

15. ## Re: First Year Uni Calculus Marathon

Originally Posted by porcupinetree
$\lim_{x \rightarrow 0} \left(\frac{1+\tan{x}}{1+\sin{x}}\right)^{\frac{1} {\sin{x}}}$
$\lim _{ x\rightarrow 0 } \left( \frac { 1+\tan { x } }{ 1+\sin { x } } \right) ^{ \frac { 1 }{ \sin { x } } }\\ =\lim _{ x\rightarrow 0 }{ { e }^{ \ln { { \left( \frac { 1+\tan { x } }{ 1+\sin { x } } \right) } } ^{ \csc { x } } } } \\ =\lim _{ x\rightarrow 0 }{ { e }^{ \frac { \ln { \left( 1+\tan { x } \right) } -\ln { \left( 1+\sin { x } \right) } }{ \sin { x } } } } \\ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+\tan { x } \right) } -\ln { \left( 1+\sin { x } \right) } }{ \sin { x } } } }\\ \stackrel{L'H}{=} { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { \sec ^{ 2 }{ x } }{ 1+\tan { x } } -\frac { \cos { x } }{ 1+\sin { x } } }{ \cos { x } } } }\\ ={ e }^{ \frac { 0 }{ 1 } }\\ =1$

$Use of L'H was justified as the final limit exists.$

Tightly squeezed LaTeX 101

16. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
$\noindent\lim _{ x\rightarrow 0 } \left( \frac { 1+\tan { x } }{ 1+\sin { x } } \right) ^{ \frac { 1 }{ \sin { x } } }\\ =\lim _{ x\rightarrow 0 }{ { e }^{ \ln { { \left( \frac { 1+\tan { x } }{ 1+\sin { x } } \right) } } ^{ \csc { x } } } } \\ =\lim _{ x\rightarrow 0 }{ { e }^{ \frac { \ln { \left( 1+\tan { x } \right) } -\ln { \left( 1+\sin { x } \right) } }{ \sin { x } } } } \\ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+\tan { x } \right) } -\ln { \left( 1+\sin { x } \right) } }{ \sin { x } } } }\\ \stackrel{L'H}{=} { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { \sec ^{ 2 }{ x } }{ 1+\tan { x } } -\frac { \cos { x } }{ 1+\sin { x } } }{ \cos { x } } } }\\ ={ e }^{ \frac { 0 }{ 1 } }\\ =1$

$Use of L'H was justified as the final limit exists.$

Tightly squeezed LaTeX 101
You moved the limit inside without justification though...

17. ## Re: First Year Uni Calculus Marathon

Lol sorry, forgot to say the exponential function was continuous even though I had it written down.

18. ## Re: First Year Uni Calculus Marathon

Originally Posted by leehuan
Lol sorry, forgot to say the exponential function was continuous even though I had it written down.
I don't recall that continuity is sufficient to exchange limits.

19. ## Re: First Year Uni Calculus Marathon

I don't recall that continuity is sufficient to exchange limits.
I'm pretty sure it's exactly one of the (equivalent) definitions of continuity?

20. ## Re: First Year Uni Calculus Marathon

Yeah you can move the limit inside for a continuous function. If lim as x -> a of g(x) is b and f is continuous at b, then lim as x -> a of f(g(x)) = f(b).

21. ## Re: First Year Uni Calculus Marathon

This is not a calculus question, but it's associated so I'll put it here.

$\noindent \alpha,\,\beta are in the first quadrant, with \alpha\geq\beta. \\\\Show that, if \\\\ \frac{\sin{\alpha}}{\sin{\beta}}\leq 1+\epsilon\\\\ then it follows that \\\\ \frac{\alpha}{\beta} \leq 1+\sqrt{\epsilon}$

22. ## Re: First Year Uni Calculus Marathon

This is not a calculus question, but it's associated so I'll put it here.

$\noindent \alpha,\,\beta are in the first quadrant, with \alpha\geq\beta. \\\\Show that, if \\\\ \frac{\sin{\alpha}}{\sin{\beta}}\leq 1+\epsilon\\\\ then it follows that \\\\ \frac{\alpha}{\beta} \leq 1+\sqrt{\epsilon}$
This is only true for sufficiently small epsilon.

23. ## Re: First Year Uni Calculus Marathon

Originally Posted by seanieg89
This is only true for sufficiently small epsilon.
Tis what I was thinking, but what is sufficient?

24. ## Re: First Year Uni Calculus Marathon

Tis what I was thinking, but what is sufficient?
Anything smaller than the first positive solution to

$\sec\left(\frac{\pi\epsilon}{2(1+\epsilon)}\right) =1+\epsilon^2.$

25. ## Re: First Year Uni Calculus Marathon

$A1. Suppose f and g are differentiable functions on the interval [a,b] such that g(a)\neq g(b). Prove that there exists c\in (a,b) such that\\ \\ \frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.\\ \\ Hint: The main ingredient is applying the extreme value theorem to a suitably chosen function. (Applying the extreme value theorem in this way is actually how one proves a different well-known theorem which can also be used in this question in place of the extreme value theorem).\\ \\A2. (Openended) Using (1) and simpler results (standard theorems from differential calculus but nothing to do with integration), state and prove whatever versions of L'Hopital's rule that you can.$

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