TD 3

Index

Limits - By Direct Computation

Exercise 1a

$$\begin{aligned} &\lim_{x \to \infty} \frac{3x^3 - 2x + 1}{x^2 - 1} \\ = \frac{x^{-2}}{x^{-2}} &\lim_{x \to \infty} \frac{3x^3 - 2x + 1}{x^2 - 1} \\ = &\lim_{x \to \infty} \frac{3x - \frac{2}{x} + \frac{1}{x^2}}{1 - \frac{1}{x^2}} \\ = &\frac{\lim_{x \to \infty} 3x - \frac{2}{x} + \frac{1}{x^2}}{\lim_{x \to \infty} 1 - \frac{1}{x^2}} \\ = &\frac{\infty - 0 + 0}{1 - 0} \\ = &\infty \\ \end{aligned}$$

Exercise 1b

$$\begin{aligned} &\lim_{x \to \infty} e^{x^2} - e^x \\ = &\lim_{x \to \infty} e^{x^2} (1 - e^{x-x^2}) \\ = &(\lim_{x \to \infty} e^{x^2}) (\lim_{x \to \infty} 1 - e^{x-x^2}) \\ = &(\infty) (1-0) \\ = &\infty \\ \end{aligned}$$

Limits - By Function Speeds

Exercise 2b

$$\begin{aligned} &\lim_{x \to -\infty} x^n e^x && \text{ with } n \in \mathbb{N} \\ = &\lim_{x \to +\infty} (-x)^n e^{-x} && \\ = &\lim_{x \to +\infty} \frac{(-x)^n}{e^x} && \\ = &0 && \text{ because } e^x > (-x)^n \\ \end{aligned}$$

Exercise 2d

$$\begin{aligned} &\lim_{x \to 0^+} \sqrt{x} ln(x) && \text{ replace } y = \frac{1}{x} \\ = &\lim_{y \to +\infty} \frac{ln(\frac{1}{y})}{\sqrt{y}} && \\ = &0 && \text{ because } y^{\frac{1}{2}} > ln(y^{-1}) \\ \end{aligned}$$

Limits - By Equivalence Relations

Exercise 3a

$$\begin{aligned} &\lim_{x \to 0^+} \frac{sin(2x^2)}{x} \\ \sim &\lim_{x \to 0^+} \frac{2x^2}{x} \\ = &\lim_{x \to 0^+} 2x \\ = &0 \\ \end{aligned}$$

Exercise 3c

$$\begin{aligned} &\lim_{x \to 0^+} \frac{cos(4x) - 1}{e^{x^2}-1} \\ \sim &\lim_{x \to 0^+} \frac{-\frac{1}{2}(4x)^2}{x^2} \\ = &\lim_{x \to 0^+} \frac{-\frac{1}{2}16x^2}{x^2} \\ = &-8 \\ \end{aligned}$$

Derivatives - By Definition

Exercise 4b

$$\begin{aligned} f(x) &= sin(x) \\ \frac{df(x)}{dx} &= \lim_{h \to 0} \frac{sin(x+h) - sin(x)}{h} \\ &= \lim_{h \to 0} \frac{sin(x)cos(h) + cos(x)sin(h) - sin(x)}{h} \\ &= \lim_{h \to 0} \frac{sin(x)(cos(h)-1)}{h} + \lim_{h \to 0} \frac{cos(x)sin(h)}{h} \\ &= \lim_{h \to 0} \frac{(cos(h)-1)}{h} sin(x) + \lim_{h \to 0} \frac{sin(h)}{h} cos(x) \\ &\sim \lim_{h \to 0} \frac{-\frac{1}{2}h^2}{h} sin(x) + \lim_{h \to 0} \frac{h}{h} cos(x) \\ &= \lim_{h \to 0} \frac{-h}{2}sin(x) + cos(x) \\ &= cos(x) \\ \end{aligned}$$

Derivatives - By Rules

Exercise 5b

$$\begin{aligned} \frac{d}{dx} [ln(\sqrt{5x^2 - 4})] &= \frac{1}{\sqrt{5x^2 - 4}} \frac{d}{dx} [(5x^2 - 4)^{\frac{1}{2}}] \\ &= (5x^2-4)^{-\frac{1}{2}} \frac{1}{2} (5x^2-4)^{-\frac{1}{2}} \frac{d}{dx} [5x^2 - 4] \\ &= \frac{1}{2} (5x^2-4)^{-1} 10x \\ &= \frac{5x}{5x^2-4} \\ \end{aligned}$$

Exercise 5c

$$\begin{aligned} \frac{d}{dx} [(e^{x^2+2})^2] &= 2e^{x^2+2} \frac{d}{dx} [e^{x^2+2}] \\ &= (2e^{x^2+2})(e^{x^2+2}) \frac{d}{dx} [x^2+2] \\ &= 2e^{2x^2+4} 2x \\ &= 4x e^{2x^2+4} \\ \end{aligned}$$ Note: This process can be simplified by performing $(e^{x^2+2})^2 = e^{2x^2+4}$ at the beginning.

Maxima and Minima of Functions

Given a function $f(x)$ at a critical point $x=x_0$, there are three cases where $f'(x_0) = 0$:

  1. If $f'(x_0)$ is increasing at $x_0$, i.e. $f''(x_0) > 0$, then $f(x)$ has a local minimum at $x_0$.
  2. If $f'(x_0)$ is decreasing at $x_0$, i.e. $f''(x_0) < 0$, then $f(x)$ has a local maximum at $x_0$.
  3. If $f'(x_0)$ is stationary at $x_0$, i.e. $f''(x_0) = 0$, then $f(x)$ has a plateau at $x_0$, but it isn't necessarily a maximum or minimum.

Exercise 6a

$$\begin{aligned} \frac{d}{dx} [x^3-3x^2-9x+12] &= 0 \\ 3x^2-6x-9 &= 0 \\ x^2-2x-3 &= 0 \\ (x-3)(x+1) &= 0 \\ x &\in \{-1,3\} \\ \end{aligned}$$ Now take the second derivative $\frac{d}{dx} [x^2-2x-3] = 2x-2$ and evaluate it with said values of $x$. $$\begin{aligned} 2(-1) - 2 &= -4 &< 0 && \text{ so there is a maximum at } x &= -1 \\ 2(3) - 2 &= 4 &> 0 && \text{ so there is a minimum at } x &= 3 \\ \end{aligned}$$

Exercise 6b

$$\begin{aligned} \frac{d}{dx} [4 \sqrt{x} - x^2] &= 0 \\ 4 \frac{1}{2\sqrt{x}} - 2x &= 0 \\ \frac{2}{\sqrt{x}} &= 2x \\ x\sqrt{x} &= 1 \\ x^3 &= 1 \\ x &\in \{1\} \\ \end{aligned}$$ Now take the second derivative $\frac{d}{dx} [2x^{-\frac{1}{2}} - 2x] = -x^{-\frac{3}{2}}-2$ and evaluate it with said value of $x$. $$\begin{aligned} -(1)^{-\frac{3}{2}}-2 &= -3 &< 0 && \text{ so there is a maximum at } x = 1 \\ \end{aligned}$$

Exercise 6c

$$\begin{aligned} \frac{d}{dx} [x + sin(x)] &= 0 && \text{ with } x \in [0, 2\pi] \\ 1+cos(x) &= 0 && \\ cos(x) &= -1 \\ x &\in {\pi} \\ \end{aligned}$$ Now take the second derivative $\frac{d}{dx} [cos(x) + 1] = -sin(x)$ and evaluate it with said value of $x$. $$-sin(\pi) = 0$$ In this case, the function $f(x) = x + sin(x)$ has a plateau at $x = \pi$ that is neither a local minimum or maximum. This is inferred by the fact that $cos(x) + 1$ is always positive (or zero), meaning that it never has the change of sign that would occur in a minimum/maximum.