Calculus & Analysis Marathon & Questions (1 Viewer)

Paradoxica · Apr 25, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
$$\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...

leehuan · Apr 25, 2016

Re: First Year Uni Calculus Marathon

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

Paradoxica · Apr 25, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
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.

KingOfActing · Apr 25, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
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?

InteGrand · Apr 25, 2016

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).

Paradoxica · Apr 27, 2016

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 $\\\\ \frac{\alpha}{\beta} \leq 1+\sqrt{\epsilon}$

seanieg89 · Apr 28, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
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.

Paradoxica · Apr 28, 2016

Re: First Year Uni Calculus Marathon

seanieg89 said:
This is only true for sufficiently small epsilon.

Tis what I was thinking, but what is sufficient?

seanieg89 · Apr 28, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
Tis what I was thinking, but what is sufficient?

Anything smaller than the first positive solution to

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

It's about 0.117041.

seanieg89 · Apr 29, 2016

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.$

porcupinetree · Apr 29, 2016

Re: First Year Uni Calculus Marathon

seanieg89 said:
$1. 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).\\ \\2. (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.$

Do you have any hints as to what the 'suitably chosen function' is? I'm struggling to find it

seanieg89 · Apr 29, 2016

Re: First Year Uni Calculus Marathon

porcupinetree said:
Do you have any hints as to what the 'suitably chosen function' is? I'm struggling to find it

It is slightly tricky to spot, but the core idea is the same as in the proof of the mean value theorem. So perhaps I should make this a leadup exercise.

The mean value theorem asserts we can find a c in (a,b) with f'(c)=(f(b)-f(a))/(b-a) if f is a differentiable function on [a,b].

B1. (Rolle's Theorem) Prove that if g is a differentiable function on [a,b] with g(a)=g(b), then g'(c)=0 for some c in (a,b). (Use the extreme value theorem to do this)

B2. (MVT) By choosing g appropriately show that f'(c)=(f(b)-f(a))/(b-a) for some c in (a,b). (The appropriate choice is easier to see here than in my original question, I think you are capable of finding it. We can also interpret this statement geometrically, which might help).

Then attempt the questions in my previous post. Aim to use Rolle's in a similar way to how it is used in Q2 of this post. With this target in mind, the choice of function should be easier for you to find.

seanieg89 · Apr 29, 2016

Re: First Year Uni Calculus Marathon

Note: To be clear on nomenclature, when I say a function is differentiable on [a,b], I am being lazy. I actually mean that f is a continuous function on [a,b] that is differentiable on (a,b). No assumptions are made about the existence of one-sided derivatives or anything at the boundary of the interval. I doubt this will affect the way anyone approaches the problem though.

leehuan · Apr 30, 2016

Re: First Year Uni Calculus Marathon

I kinda want to make this thread more accessible to some people (where possible) so here's a somewhat easy question. If it's ignored too bad back to reasonable difficulty.

$\\$By considering the function $g:[a,b]\rightarrow \mathbb R, g(x) = f(x) - \left(\frac{f(b)-f(a)}{b-a}(x-a)+f(a) \right)$ where $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, prove the statement of the mean value theorem.$$

Notes:

Some other theorem should be assumed.

InteGrand · Apr 30, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
I kinda want to make this thread more accessible to some people (where possible) so here's a somewhat easy question. If it's ignored too bad back to reasonable difficulty.

$\\$By considering the function $g:[a,b]\rightarrow \mathbb R, g(x) = f(x) - \left(\frac{f(b)-f(a)}{b-a}(x-a)+f(a) \right)$ where $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, prove the statement of the mean value theorem.$$

Notes:

Some other theorem should be assumed.

This is basically a hint to one of seanieg89's hint exercises.

leehuan · Apr 30, 2016

Re: First Year Uni Calculus Marathon

InteGrand said:
This is basically a hint to one of seanieg89's hint exercises.

I, did not see that coming.

seanieg89 · Apr 30, 2016

Re: First Year Uni Calculus Marathon

It's just providing the function for my question B2 (2nd question in second post) which is not too much of a spoiler, but now someone should definitely be able to prove that.

B1 is a bit trickier, but for any student who wants to assume Rolle's and doesn't care where it comes from, you can move straight on to the questions in my first post which was the main point (to prove L'Hopital's).

seanieg89 · Apr 30, 2016

Re: First Year Uni Calculus Marathon

And some much easier ones for students who don't want to do the above questions:

E1. Prove that if a function f: (a,b) -> R is differentiable and f'(x) is non-negative in this interval, then f is non-decreasing in this interval.

E2. Prove that if a function f: (a,b) -> R is differentiable and f'(x) = 0 in this interval, then f is constant.

leehuan · Apr 30, 2016

Re: First Year Uni Calculus Marathon

I thought E2 was kinda intuitive but I'm finding it hard to word my final bit.

Rolle's theorem dictates that if f is continuous on [a,b] and differentiable on (a,b), and we have f(a)=f(b), then there exists at least one value of c such that f'(c)=0 for c in (a,b)

But if f'(c)=0 for all (a,b) (or alternatively f has a horizontal tangent for all points on the interval), then as we have an infinite number of values satisfying f'(c)=0, f must be constant for every c in [a,b]

seanieg89 · Apr 30, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
But if f'(c)=0 for all (a,b) (or alternatively f has a horizontal tangent for all points on the interval), then as we have an infinite number of values satisfying f'(c)=0, f must be constant for every c in [a,b]

What do you mean by saying f must be constant for every c? What does it mean to be constant at a point?

Calculus & Analysis Marathon & Questions (1 Viewer)

-insert title here-

Well-Known Member

-insert title here-

lukewarm mess

Well-Known Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

not actually a porcupine

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)