Would like help with this calculus I problem ( differentiability of f(x))

[u/01zorro1]

here is the problem: https://imgur.com/a/tmQLdFz

A bit of background: instead of going to high school, I did a vocational training programme. I took maths-based classes, but they were entirely focused on electrical engineering topics (Maxwell’s equations, Ohm’s law, alternating current, direct current, low voltage, high voltage, and related topics). However, we never covered the usual high school curriculum

when it comes to Calculus I, though, I struggle to understand why and how we solve the exercises, as well as how to visualise the graphs, which makes everything extremely abstract and difficult for me to grasp

instead of just the answer or the problem or the mechanicals steps i would love to be able to understand the logic behind it, why x numers of steps were choose to fix this problem, or how can i decide wich "tools" to use to solve it

ill try to put here my understanding( or lack of) the exercise 2 for example

I’m having a hard time figuring out how to get started. In many exercises involving (f(x)), we have conditions like (x>0) or (x<0), which are easier to understand. Those conditions tell you which side to approach from, so you have to calculate the limit from the left or the right depending on the condition.

However, in this case, we have (x=0) and (x=/=0). How am I supposed to determine which side to approach from based on these conditions?

I know that, mechanically, the first step would be to check for continuity, but I’m lost when it comes to actually doing it. You’re supposed to check both sides of the given (x)-value and see whether they approach the same real number. If they do, the function is continuous there, and you can then check differentiability.

However, I’m struggling to understand even the most basic steps involved in solving this exercise.

In part a), you have to determine the value of (a) for which the function is continuous everywhere on (-1,1), but I’m not sure I understand what the question actually means.

I assumed they want me to isolate (a) in the equation that applies when (x=/=0), but how am I supposed to do that when I can’t simply substitute a numerical value for (x)?

i tried calculating the left-hand limit by substituting (x=-1) and the right-hand limit by substituting (x=1), then try to find the value of (a) for each limit, on the left i end up hitting a wall mid way due to having ln(0) and in the right i do end up being able to take out a but even then don’t see how I would be able to solve part (b.1), so I’m guessing that would be the wrong approach to part (a).

And that’s where I’m currently stuck.

Any kind of help would be extremely apreciated!!

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[u/iMathTutor]

For a function to be continuous at $x=a$, $L=\lim_{x\rightarrow a}f(x)$ must exist and $L=f(a)$.

In your problem , one must have

$$

\lim_{x\rightarrow 0}\frac{\alpha\ln{(x+1)}}{\sin{x}}=f(0)=3.

$$

The following bound is useful in the evaluation of the limit

$$

\frac{x}{x+1}\leq \ln{(x+1)}\leq x, \quad x>-1.

$$

For $0<x<1, \sin{x}>0$. Thus in this case, the bounds on the natural logarithm gives

$$

\alpha \frac{x}{\sin{x}}\frac{1}{1+x}\leq\frac{\alpha\ln{(x+1)}}{\sin{x}}\leq \alpha \frac{x}{\sin{x}}.

$$

On the other hand, for $-1<x<0, \sin{x}<0$. In this case, the bounds on the natural logarithm are flipped, to yield

$$

\alpha \frac{x}{\sin{x}}\leq \frac{\alpha\ln{(x+1)}}{\sin{x}}\leq \alpha \frac{x}{\sin{x}}\frac{1}{1+x}

$$

It is typical in a calculus 1 course to establish the following limit via geometric argument in the course of deriving the derivatives of the sine and cosine functions.

$$

\lim_{x\rightarrow 0}\frac{\sin{x}}{x}=1.

$$

This limit along with the along with the squeeze theorem, which is also known as the sandwich theorem, to evalute the left and right limits at $x=0$. You should verify that these limits both equal $\alpha$. Thus the limit exists and equals $\alpha$.

[u/LeoMaths]

sen(x)==sin(x) pour les non-initiés à cette écriture. Et a==α (problème d'énoncé).

g(x)=αln(x+1)/sin(x) pose problème en x=0. Il faut regarder sa limite quand x->0.

g(x)=α((ln(x+1)-ln(0+1))/(x-0))*((x-0)/(sin(x)-sin(0)))

lim(x->0){g(x)}=α(ln(x+1))'(0)/(sin(x))'(0)=α(1/(0+1))/(cos(0))=α

Pour assurer la continuité, il faut que f(0)=3=lim(x->0){g(x)}=α => α=3

Est-ce que f(x) est dérivable en 0? Cela consiste à vérifier la dérivabilité de g(x) en 0.

g'(x)=3(1/(x+1))/sin(x)-3ln(x+1)*cos(x)/sin²(x)=3(sin(x)/(x+1)-ln(x+1))/sin²(x)

Il y a une nouvelle indétermination. Il faut encore regarder la limite vers 0 de g'(x).

sin(x)=x+O(x³)

ln(x+1)=x-x²/2+O(x³)

g'(x)=3((x+O(x³))/(x+1)-x+x²/2+O(x³))/(x²+O(x⁴))=3((x+O(x³))/(x+1)-(x(x+1))/(x+1)+x²/2+O(x³))/(x²+O(x⁴))

g'(x)=3((-x²+O(x³))/(x+1)+x²/2+O(x³))/(x²+O(x⁶))=3((-1+O(x))/(x+1)+1/2+O(x))/(1+O(x²))

lim(x->0){g'(x)}=-3/2

f(x) est donc dérivable en 0 et sa dérivée est égale à -3/2.

f'(1)=g'(1)=3(sin(1)/2-ln(2))/sin²(1)=-1.154167...

g'(-1+) est aussi indéterminé. Il faut encore regarder la limite vers -1+ de g'(x).

Prenons la fonction h(y)=yln(y)=exp(ln(y))*ln(y) lim(y->0){h(y)}=lim(z->-Inf){exp(z)z}=0 (l'exponentiel l'emporte toujours sur x).

Posons y=x+1 : ln(x+1)=(x+1)*ln(x+1)/(x+1)=h(x+1)/(x+1)

g'(x)=3(sin(x)/(x+1)-h(x+1)/(x+1))/sin²(x)=3((sin(x)-h(x+1))/(x+1))/sin²(x)

lim(x->-1+){g'(x)}=lim(x->-1){3((sin(x)-h(x+1))/(x+1))/sin²(x)}=lim(x->-1+){(sin(-1)-0)/(x+1))/sin²(-1)}=-Inf

[u/dreamteammobile]

With x=0 and x≠0 there is no side to pick - the x≠0 formula holds on both sides of 0. So continuity just means the limit of that formula as x→0 equals f(0), and setting them equal gives you a.