Solved on Feb 09, 2024

Exercise 1: Limits 3) Write the reference limits of the function $\ln x$ 4) Using these limits, calculate the limits of the following functions at the bounds of their domain: a) $f(x)=x^{2}-2+\ln x$ b) $g(x)=x^{2}+(2-\ln x)^{2}$ c) $h(x)=\frac{\ln (x+2)}{1+x^{2}}$ d) $l(x)=x(\ln x)^{2}$
Exercise 2: Derivative and sign study of the derivative Calculate the derivative function of the following functions and study the sign of the derivative: 4) $f(x)=-1+(\ln x)^{2}$ 5) $g(x)=\ln \frac{x}{x+1}$ 6) $h(x)=\ln \left|\frac{x-2}{x+4}\right|$
Exercise 3: Complex numbers 3) Solve the equation $z^{2}-2 i z-2=0$ in the set $\mathbb{C}$ of complex numbers 4) Let $z_{1}$ and $z_{2}$ be the solutions of this equation such that $\operatorname{Re}\left(z_{1}\right)>\operatorname{Re}\left(z_{2}\right)$ .

STEP 1

Exercice 1 : Limites
Assumptions
1. We are dealing with the natural logarithm function, denoted as $\ln$ .
2. We will use the standard limits of the natural logarithm function to solve the problems.
3. The domain of the natural logarithm function is $(0, +\infty)$ .

STEP 2

Write the reference limits for the natural logarithm function.
$\lim_{x \to 0^+} \ln(x) = -\infty$ $\lim_{x \to +\infty} \ln(x) = +\infty$

STEP 3

Calculate the limit of $f(x) = x^2 - 2 + \ln(x)$ as $x$ approaches $0$ from the right.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 - 2 + \ln(x))$

STEP 4

Since $\lim_{x \to 0^+} \ln(x) = -\infty$ and $\lim_{x \to 0^+} x^2 - 2$ is finite, the dominant term is $\ln(x)$ .
$\lim_{x \to 0^+} f(x) = -\infty$

STEP 5

Calculate the limit of $f(x) = x^2 - 2 + \ln(x)$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (x^2 - 2 + \ln(x))$

STEP 6

As $x$ approaches $+\infty$ , both $x^2$ and $\ln(x)$ approach $+\infty$ , but $x^2$ grows much faster than $\ln(x)$ .
$\lim_{x \to +\infty} f(x) = +\infty$

STEP 7

Calculate the limit of $g(x) = x^2 + (2 - \ln(x))^2$ as $x$ approaches $0$ from the right.
$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} \left(x^2 + (2 - \ln(x))^2\right)$

STEP 8

Since $\lim_{x \to 0^+} \ln(x) = -\infty$ , the term $(2 - \ln(x))^2$ will dominate and approach $+\infty$ .
$\lim_{x \to 0^+} g(x) = +\infty$

STEP 9

Calculate the limit of $g(x) = x^2 + (2 - \ln(x))^2$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} g(x) = \lim_{x \to +\infty} \left(x^2 + (2 - \ln(x))^2\right)$

STEP 10

As $x$ approaches $+\infty$ , $x^2$ grows much faster than $\ln(x)$ , and the term $(2 - \ln(x))^2$ approaches a constant value.
$\lim_{x \to +\infty} g(x) = +\infty$

STEP 11

Calculate the limit of $h(x) = \frac{\ln(x + 2)}{1 + x^2}$ as $x$ approaches $-\infty$ .
$\lim_{x \to -\infty} h(x) = \lim_{x \to -\infty} \frac{\ln(x + 2)}{1 + x^2}$

STEP 12

As $x$ approaches $-\infty$ , $x + 2$ approaches $-\infty$ , but since the natural logarithm is not defined for negative numbers, this limit does not exist.
$\lim_{x \to -\infty} h(x) \text{ does not exist}$

STEP 13

Calculate the limit of $h(x) = \frac{\ln(x + 2)}{1 + x^2}$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} h(x) = \lim_{x \to +\infty} \frac{\ln(x + 2)}{1 + x^2}$

STEP 14

As $x$ approaches $+\infty$ , the denominator $1 + x^2$ grows much faster than the numerator $\ln(x + 2)$ , leading to the limit being $0$ .
$\lim_{x \to +\infty} h(x) = 0$

STEP 15

Calculate the limit of $l(x) = x(\ln x)^2$ as $x$ approaches $0$ from the right.
$\lim_{x \to 0^+} l(x) = \lim_{x \to 0^+} x(\ln x)^2$

STEP 16

As $x$ approaches $0$ from the right, $\ln(x)$ approaches $-\infty$ , but since $x$ approaches $0$ , the product $x(\ln x)^2$ approaches $0$ .
$\lim_{x \to 0^+} l(x) = 0$

STEP 17

Calculate the limit of $l(x) = x(\ln x)^2$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} l(x) = \lim_{x \to +\infty} x(\ln x)^2$

STEP 18

As $x$ approaches $+\infty$ , both $x$ and $(\ln x)^2$ approach $+\infty$ , so the limit is $+\infty$ .
$\lim_{x \to +\infty} l(x) = +\infty$

Was this helpful?

STEP 1

Exercice 1 : Limites
Assumptions
1. We are dealing with the natural logarithm function, denoted as $\ln$ .
2. We will use the standard limits of the natural logarithm function to solve the problems.
3. The domain of the natural logarithm function is $(0, +\infty)$ .

STEP 2

Write the reference limits for the natural logarithm function.
$\lim_{x \to 0^+} \ln(x) = -\infty$ $\lim_{x \to +\infty} \ln(x) = +\infty$

STEP 3

Calculate the limit of $f(x) = x^2 - 2 + \ln(x)$ as $x$ approaches $0$ from the right.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 - 2 + \ln(x))$

STEP 4

Since $\lim_{x \to 0^+} \ln(x) = -\infty$ and $\lim_{x \to 0^+} x^2 - 2$ is finite, the dominant term is $\ln(x)$ .
$\lim_{x \to 0^+} f(x) = -\infty$

STEP 5

Calculate the limit of $f(x) = x^2 - 2 + \ln(x)$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (x^2 - 2 + \ln(x))$

STEP 6

As $x$ approaches $+\infty$ , both $x^2$ and $\ln(x)$ approach $+\infty$ , but $x^2$ grows much faster than $\ln(x)$ .
$\lim_{x \to +\infty} f(x) = +\infty$

STEP 7

Calculate the limit of $g(x) = x^2 + (2 - \ln(x))^2$ as $x$ approaches $0$ from the right.
$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} \left(x^2 + (2 - \ln(x))^2\right)$

STEP 8

Since $\lim_{x \to 0^+} \ln(x) = -\infty$ , the term $(2 - \ln(x))^2$ will dominate and approach $+\infty$ .
$\lim_{x \to 0^+} g(x) = +\infty$

STEP 9

Calculate the limit of $g(x) = x^2 + (2 - \ln(x))^2$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} g(x) = \lim_{x \to +\infty} \left(x^2 + (2 - \ln(x))^2\right)$

STEP 10

As $x$ approaches $+\infty$ , $x^2$ grows much faster than $\ln(x)$ , and the term $(2 - \ln(x))^2$ approaches a constant value.
$\lim_{x \to +\infty} g(x) = +\infty$

STEP 11

Calculate the limit of $h(x) = \frac{\ln(x + 2)}{1 + x^2}$ as $x$ approaches $-\infty$ .
$\lim_{x \to -\infty} h(x) = \lim_{x \to -\infty} \frac{\ln(x + 2)}{1 + x^2}$

STEP 12

As $x$ approaches $-\infty$ , $x + 2$ approaches $-\infty$ , but since the natural logarithm is not defined for negative numbers, this limit does not exist.
$\lim_{x \to -\infty} h(x) \text{ does not exist}$

STEP 13

Calculate the limit of $h(x) = \frac{\ln(x + 2)}{1 + x^2}$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} h(x) = \lim_{x \to +\infty} \frac{\ln(x + 2)}{1 + x^2}$

STEP 14

As $x$ approaches $+\infty$ , the denominator $1 + x^2$ grows much faster than the numerator $\ln(x + 2)$ , leading to the limit being $0$ .
$\lim_{x \to +\infty} h(x) = 0$

STEP 15

Calculate the limit of $l(x) = x(\ln x)^2$ as $x$ approaches $0$ from the right.
$\lim_{x \to 0^+} l(x) = \lim_{x \to 0^+} x(\ln x)^2$

STEP 16

As $x$ approaches $0$ from the right, $\ln(x)$ approaches $-\infty$ , but since $x$ approaches $0$ , the product $x(\ln x)^2$ approaches $0$ .
$\lim_{x \to 0^+} l(x) = 0$

STEP 17

Calculate the limit of $l(x) = x(\ln x)^2$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} l(x) = \lim_{x \to +\infty} x(\ln x)^2$

STEP 18

As $x$ approaches $+\infty$ , both $x$ and $(\ln x)^2$ approach $+\infty$ , so the limit is $+\infty$ .
$\lim_{x \to +\infty} l(x) = +\infty$

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

STEP 1

Exercice 1 : LimitesAssumptions1. We are dealing with the natural logarithm function, denoted as ln⁡\lnln.2. We will use the standard limits of the natural logarithm function to solve the problems.3. The domain of the natural logarithm function is (0,+∞)(0, +\infty)(0,+∞).

STEP 2

Write the reference limits for the natural logarithm function.lim⁡x→0+ln⁡(x)=−∞\lim_{x \to 0^+} \ln(x) = -\inftyx→0+lim​ln(x)=−∞ lim⁡x→+∞ln⁡(x)=+∞\lim_{x \to +\infty} \ln(x) = +\inftyx→+∞lim​ln(x)=+∞

STEP 3

Calculate the limit of f(x)=x2−2+ln⁡(x)f(x) = x^2 - 2 + \ln(x)f(x)=x2−2+ln(x) as xxx approaches 000 from the right.lim⁡x→0+f(x)=lim⁡x→0+(x2−2+ln⁡(x))\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 - 2 + \ln(x))x→0+lim​f(x)=x→0+lim​(x2−2+ln(x))

STEP 4

Since lim⁡x→0+ln⁡(x)=−∞\lim_{x \to 0^+} \ln(x) = -\inftylimx→0+​ln(x)=−∞ and lim⁡x→0+x2−2\lim_{x \to 0^+} x^2 - 2limx→0+​x2−2 is finite, the dominant term is ln⁡(x)\ln(x)ln(x).lim⁡x→0+f(x)=−∞\lim_{x \to 0^+} f(x) = -\inftyx→0+lim​f(x)=−∞

STEP 5

Calculate the limit of f(x)=x2−2+ln⁡(x)f(x) = x^2 - 2 + \ln(x)f(x)=x2−2+ln(x) as xxx approaches +∞+\infty+∞.lim⁡x→+∞f(x)=lim⁡x→+∞(x2−2+ln⁡(x))\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (x^2 - 2 + \ln(x))x→+∞lim​f(x)=x→+∞lim​(x2−2+ln(x))

STEP 6

As xxx approaches +∞+\infty+∞, both x2x^2x2 and ln⁡(x)\ln(x)ln(x) approach +∞+\infty+∞, but x2x^2x2 grows much faster than ln⁡(x)\ln(x)ln(x).lim⁡x→+∞f(x)=+∞\lim_{x \to +\infty} f(x) = +\inftyx→+∞lim​f(x)=+∞

STEP 7

STEP 8

Since lim⁡x→0+ln⁡(x)=−∞\lim_{x \to 0^+} \ln(x) = -\inftylimx→0+​ln(x)=−∞, the term (2−ln⁡(x))2(2 - \ln(x))^2(2−ln(x))2 will dominate and approach +∞+\infty+∞.lim⁡x→0+g(x)=+∞\lim_{x \to 0^+} g(x) = +\inftyx→0+lim​g(x)=+∞

STEP 9

STEP 10

As xxx approaches +∞+\infty+∞, x2x^2x2 grows much faster than ln⁡(x)\ln(x)ln(x), and the term (2−ln⁡(x))2(2 - \ln(x))^2(2−ln(x))2 approaches a constant value.lim⁡x→+∞g(x)=+∞\lim_{x \to +\infty} g(x) = +\inftyx→+∞lim​g(x)=+∞

STEP 11

STEP 12

As xxx approaches −∞-\infty−∞, x+2x + 2x+2 approaches −∞-\infty−∞, but since the natural logarithm is not defined for negative numbers, this limit does not exist.lim⁡x→−∞h(x) does not exist\lim_{x \to -\infty} h(x) \text{ does not exist}x→−∞lim​h(x) does not exist

STEP 13

Calculate the limit of h(x)=ln⁡(x+2)1+x2h(x) = \frac{\ln(x + 2)}{1 + x^2}h(x)=1+x2ln(x+2)​ as xxx approaches +∞+\infty+∞.lim⁡x→+∞h(x)=lim⁡x→+∞ln⁡(x+2)1+x2\lim_{x \to +\infty} h(x) = \lim_{x \to +\infty} \frac{\ln(x + 2)}{1 + x^2}x→+∞lim​h(x)=x→+∞lim​1+x2ln(x+2)​

STEP 14

As xxx approaches +∞+\infty+∞, the denominator 1+x21 + x^21+x2 grows much faster than the numerator ln⁡(x+2)\ln(x + 2)ln(x+2), leading to the limit being 000.lim⁡x→+∞h(x)=0\lim_{x \to +\infty} h(x) = 0x→+∞lim​h(x)=0

STEP 15

Calculate the limit of l(x)=x(ln⁡x)2l(x) = x(\ln x)^2l(x)=x(lnx)2 as xxx approaches 000 from the right.lim⁡x→0+l(x)=lim⁡x→0+x(ln⁡x)2\lim_{x \to 0^+} l(x) = \lim_{x \to 0^+} x(\ln x)^2x→0+lim​l(x)=x→0+lim​x(lnx)2

STEP 16

As xxx approaches 000 from the right, ln⁡(x)\ln(x)ln(x) approaches −∞-\infty−∞, but since xxx approaches 000, the product x(ln⁡x)2x(\ln x)^2x(lnx)2 approaches 000.lim⁡x→0+l(x)=0\lim_{x \to 0^+} l(x) = 0x→0+lim​l(x)=0

STEP 17

Calculate the limit of l(x)=x(ln⁡x)2l(x) = x(\ln x)^2l(x)=x(lnx)2 as xxx approaches +∞+\infty+∞.lim⁡x→+∞l(x)=lim⁡x→+∞x(ln⁡x)2\lim_{x \to +\infty} l(x) = \lim_{x \to +\infty} x(\ln x)^2x→+∞lim​l(x)=x→+∞lim​x(lnx)2

STEP 18

As xxx approaches +∞+\infty+∞, both xxx and (ln⁡x)2(\ln x)^2(lnx)2 approach +∞+\infty+∞, so the limit is +∞+\infty+∞.lim⁡x→+∞l(x)=+∞\lim_{x \to +\infty} l(x) = +\inftyx→+∞lim​l(x)=+∞

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Exercice 1 : Limites
Assumptions
1. We are dealing with the natural logarithm function, denoted as $\ln$ .
2. We will use the standard limits of the natural logarithm function to solve the problems.
3. The domain of the natural logarithm function is $(0, +\infty)$ .

Write the reference limits for the natural logarithm function.
$\lim_{x \to 0^+} \ln(x) = -\infty$ $\lim_{x \to +\infty} \ln(x) = +\infty$

Calculate the limit of $f(x) = x^2 - 2 + \ln(x)$ as $x$ approaches $0$ from the right.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 - 2 + \ln(x))$

Since $\lim_{x \to 0^+} \ln(x) = -\infty$ and $\lim_{x \to 0^+} x^2 - 2$ is finite, the dominant term is $\ln(x)$ .
$\lim_{x \to 0^+} f(x) = -\infty$

Calculate the limit of $f(x) = x^2 - 2 + \ln(x)$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (x^2 - 2 + \ln(x))$

As $x$ approaches $+\infty$ , both $x^2$ and $\ln(x)$ approach $+\infty$ , but $x^2$ grows much faster than $\ln(x)$ .
$\lim_{x \to +\infty} f(x) = +\infty$

Since $\lim_{x \to 0^+} \ln(x) = -\infty$ , the term $(2 - \ln(x))^2$ will dominate and approach $+\infty$ .
$\lim_{x \to 0^+} g(x) = +\infty$

As $x$ approaches $+\infty$ , $x^2$ grows much faster than $\ln(x)$ , and the term $(2 - \ln(x))^2$ approaches a constant value.
$\lim_{x \to +\infty} g(x) = +\infty$

As $x$ approaches $-\infty$ , $x + 2$ approaches $-\infty$ , but since the natural logarithm is not defined for negative numbers, this limit does not exist.
$\lim_{x \to -\infty} h(x) \text{ does not exist}$

Calculate the limit of $h(x) = \frac{\ln(x + 2)}{1 + x^2}$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} h(x) = \lim_{x \to +\infty} \frac{\ln(x + 2)}{1 + x^2}$

As $x$ approaches $+\infty$ , the denominator $1 + x^2$ grows much faster than the numerator $\ln(x + 2)$ , leading to the limit being $0$ .
$\lim_{x \to +\infty} h(x) = 0$

Calculate the limit of $l(x) = x(\ln x)^2$ as $x$ approaches $0$ from the right.
$\lim_{x \to 0^+} l(x) = \lim_{x \to 0^+} x(\ln x)^2$

As $x$ approaches $0$ from the right, $\ln(x)$ approaches $-\infty$ , but since $x$ approaches $0$ , the product $x(\ln x)^2$ approaches $0$ .
$\lim_{x \to 0^+} l(x) = 0$

Calculate the limit of $l(x) = x(\ln x)^2$ as $x$ approaches $+\infty$ .
$\lim_{x \to +\infty} l(x) = \lim_{x \to +\infty} x(\ln x)^2$

As $x$ approaches $+\infty$ , both $x$ and $(\ln x)^2$ approach $+\infty$ , so the limit is $+\infty$ .
$\lim_{x \to +\infty} l(x) = +\infty$