Math  /  Calculus

QuestionExercice 1. 1) Montrer en utilisant la définition que limx14x+2=6,limx0x3+x+2=2,limx+1x3+1=0\lim _{x \rightarrow 1} 4 x+2=6, \lim _{x \rightarrow 0} x^{3}+x+2=2, \lim _{x \rightarrow+\infty} \frac{1}{x^{3}+1}=0 2) Montrer que les limites limx1E(x),limx0cos1x\lim _{x \rightarrow 1} E(x), \lim _{x \rightarrow 0} \cos \frac{1}{x} n'existent pas.
Exercice 2. Calculer les limites suivantes:
1. limx0tanxsinxx3\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}},
2. limxπ2ecosx1xπ2\lim _{x \rightarrow \frac{\pi}{2}} \frac{e^{\cos x}-1}{x-\frac{\pi}{2}},
3. limx0xaE\lim _{x \rightarrow 0} \frac{x}{a} E 12x)-1-2 x),
5. limx+(x2+2x12x)\lim _{x \rightarrow+\infty}\left(\sqrt{x^{2}+2 x-1}-2 x\right), 6. limx+(1+1x)x\lim _{x \rightarrow+\infty}\left(1+\frac{1}{x}\right)^{x},
7. limx(x1x+1)x\lim _{x \rightarrow-\infty}\left(\frac{x-1}{x+1}\right)^{x}.

Exercice 3. Soit f:RRf: \mathbb{R} \longrightarrow \mathbb{R}, une fonction définie par f(x)={x si x1x2+ax+b si x>1f(x)=\left\{\begin{array}{ll} x & \text { si }|x| \leqslant 1 \\ x^{2}+a x+b & \text { si }|x|>1 \end{array}\right.
a,bRa, b \in \mathbb{R}. Déterminer les valeurs de aa et bb pour que ff soit continue sur R\mathbb{R}. Exercice 4. Soit f:RRf: \mathbb{R} \longrightarrow \mathbb{R} la fonction définie par f(x)={1 si xQ0 si Q.f(x)=\left\{\begin{array}{l} 1 \text { si } x \in \mathbb{Q} \\ 0 \text { si } \notin \mathbb{Q} . \end{array}\right.
Montrer que ff est discontinue en tout point de R\mathbb{R}. Exercice 5. Soit ff une fonction définie par f:[a,b][a,b]f:[a, b] \rightarrow[a, b], telle que f(x1)f(x2)<x1x2,x1,x2[a,b],(x1x2).\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\left|x_{1}-x_{2}\right|, \quad \forall x_{1}, x_{2} \in[a, b], \quad\left(x_{1} \neq x_{2}\right) . 1) Montrer que ff est continue sur [a,b][a, b]. 2) Montrer que l'équation f(x)=xf(x)=x admet une solution unique.
Exercice 6. Étudier dans chacun des cas suivants si la fonction ff est prolongeable par continuité sur R\mathbb{R}. 1) f:x1cosxxf: x \longrightarrow \frac{1-\cos \sqrt{|x|}}{|x|}, 2) f:xsin(x)sin(1x)f: x \longrightarrow \sin (x) \cdot \sin \left(\frac{1}{x}\right), 3) f:xx3+1x2+3x+2f: x \longrightarrow \frac{x^{3}+1}{x^{2}+3 x+2}, 4) f:x1cos(1cos1x2)f: x \longrightarrow 1-\cos \left(1-\cos \frac{1}{x-2}\right), 5) f:xsinx4x4f: x \longrightarrow \frac{\sin |x-4|}{x-4}, 6) f:x11x21x2f: x \longrightarrow \frac{1}{1-x}-\frac{2}{1-x^{2}}.

Studdy Solution

STEP 1

What is this asking? This exercise set explores limits, continuity, and some tricky function behavior, asking us to prove limits exist (or don't!), calculate limits, and figure out how to make a piecewise function continuous. Watch out! Remember the difference between proving a limit and calculating it!
Also, continuity requires checking left-hand and right-hand limits, especially with piecewise functions.

STEP 2

1. Exercise 1: Proving and Disproving Limits
2. Exercise 2: Calculating Limits
3. Exercise 3: Piecewise Continuity
4. Exercise 4: Discontinuity of a Rational Function
5. Exercise 5: Continuity and Unique Solutions
6. Exercise 6: Continuity Extensions

STEP 3

**Prove** limx1(4x+2)=6\lim_{x \rightarrow 1} (4x + 2) = 6 using the definition.
Let ϵ>0\epsilon > 0.
We want to find a δ>0\delta > 0 such that if 0<x1<δ0 < |x - 1| < \delta, then (4x+2)6<ϵ| (4x + 2) - 6 | < \epsilon.
Let's simplify that last inequality: 4x4<ϵ|4x - 4| < \epsilon, which means 4x1<ϵ4|x - 1| < \epsilon, or x1<ϵ4|x - 1| < \frac{\epsilon}{4}.
So, we can choose δ=ϵ4\delta = \frac{\epsilon}{4}.
If 0<x1<δ0 < |x - 1| < \delta, then 4x+26=4x4=4x1<4ϵ4=ϵ|4x + 2 - 6| = |4x - 4| = 4|x - 1| < 4 \cdot \frac{\epsilon}{4} = \epsilon.
Boom!

STEP 4

**Prove** limx0(x3+x+2)=2\lim_{x \rightarrow 0} (x^3 + x + 2) = 2.
Let ϵ>0\epsilon > 0.
We need to find δ>0\delta > 0 such that if 0<x0<δ0 < |x - 0| < \delta, then x3+x+22<ϵ|x^3 + x + 2 - 2| < \epsilon.
This simplifies to x3+x<ϵ|x^3 + x| < \epsilon.
Since x3+xx3+x|x^3 + x| \le |x^3| + |x|, if we can make both x3<ϵ2|x^3| < \frac{\epsilon}{2} and x<ϵ2|x| < \frac{\epsilon}{2}, we're good.
Let's choose δ=min(1,ϵ2)\delta = \min(1, \frac{\epsilon}{2}).
If x<δ|x| < \delta, then x3<x<δϵ2|x^3| < |x| < \delta \le \frac{\epsilon}{2}.
Also, x<δϵ2|x| < \delta \le \frac{\epsilon}{2}.
Therefore, x3+xx3+x<ϵ2+ϵ2=ϵ|x^3 + x| \le |x^3| + |x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.
Nailed it!

STEP 5

**Prove** limx1x3+1=0\lim_{x \rightarrow \infty} \frac{1}{x^3 + 1} = 0.
Let ϵ>0\epsilon > 0.
We want to find MM such that if x>Mx > M, then 1x3+10<ϵ|\frac{1}{x^3 + 1} - 0| < \epsilon.
This means 1x3+1<ϵ\frac{1}{x^3 + 1} < \epsilon.
Since x3+1>x3x^3 + 1 > x^3 for x>0x > 0, we have 1x3+1<1x3\frac{1}{x^3 + 1} < \frac{1}{x^3}.
If we make 1x3<ϵ\frac{1}{x^3} < \epsilon, then x3>1ϵx^3 > \frac{1}{\epsilon}, so x>1ϵ3x > \frac{1}{\sqrt[3]{\epsilon}}.
Choose M=1ϵ3M = \frac{1}{\sqrt[3]{\epsilon}}.
If x>Mx > M, then 1x3+1<1x3<1M3=ϵ\frac{1}{x^3 + 1} < \frac{1}{x^3} < \frac{1}{M^3} = \epsilon.
Perfect!

STEP 6

**Show** limx1E(x)\lim_{x \rightarrow 1} E(x) **does not exist**.
The floor function E(x)E(x) jumps at every integer.
As xx approaches 11 from the left, E(x)E(x) is 00.
As xx approaches 11 from the right, E(x)E(x) is 11.
Since the left and right limits are different, the limit does not exist.

STEP 7

**Show** limx0cos(1x)\lim_{x \rightarrow 0} \cos(\frac{1}{x}) **does not exist**.
As xx approaches 00, 1x\frac{1}{x} becomes very large, and cos(1x)\cos(\frac{1}{x}) oscillates rapidly between 1-1 and 11.
It doesn't settle down to any single value, so the limit does not exist.

STEP 8

(Presents the final answers to all exercises)

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