All my math notes, now in Markdown.

You will be able to

- Evaluate limits of composition functions
- Use Squeeze Theorem to find limits

Property | Description |
---|---|

$\lim_{x\to 0}c = c$ | The limit of a constant is equal to the constant. |

$\lim_{x\to a}x = a$ | The limit of $x$ as it approaches $a$ is equal to $a$. |

$\lim_{x\to 0}[f(x) + g(x)] = \lim_{x\to 0} f(x) + \lim_{x\to 0} g(x)$ | The limit of a sum is the sum of the limits. |

$\lim_{x\to 0}[f(x) - g(x)] = \lim_{x\to 0}f(x) - \lim_{x\to 0}g(x)$ | The limit of a difference is the difference of limits. |

$\lim_{x\to 0}[cf(x)] = c\lim_{x\to 0}f(x)$ | The limit of a constant times a function is the constant times the limit of the function. |

$\lim_{x\to 0}[f(x)g(x)] = \lim_{x\to 0}f(x)\lim_{x\to 0}g(x)$ | The limit of a product is the product of the limits. |

$\lim_{x\to 0}\frac{f(x)}{g(x)} = \frac{\lim_{x\to 0}f(x)}{\lim_{x\to 0}g(x)}$, provided $\lim_{x\to 0}g(x) = 0$ | The limit of a quotient is the quotient of the limits, provided that the denominator does not equal zero. |

$\lim_{x\to 0}[f(x)]^n = (\lim_{x\to 0}f(x))^n$, where $a$ is a rational number | The limit of a power is the power of the limit, provided that the exponent is a rational number. |

$\lim_{x\to 0}\sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to 0}f(x)}$, if the root on the right side exists | The limit of a root is the root of the limit, provided that the root exists. |

If $f$ and $g$ are functions such that $\lim_{x\to c}g(x) = L$ and $\lim_{x\to L}f(x) = f(L)$, then \(\lim_{x\to c}(f \circ g)(x) = \lim_{x\to c}f(g(x)) = f(\lim_{x\to c}g(x)) = f(L)\)

The limit of a composition is the composition of the limit, provided the outside function is continuous at the limit of the inside function.

We know by comparing rates and / or looking at the graph that $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$. How do we prove it? (Lâ€™Hospital not born yet)

If $f(x) \leq h(x) \leq g(x)$ for all $x\neq c$ in some interval about $c$, and $\lim_{x\to c}g(x) = \lim_{x\to c}f(x) = L$ then \(\lim_{x\to c}h(x) = L\)

Show the following analytically
\(\lim_{x\to 0}(x^2 \sin(1/x)) = 0\)
We know the following, $-x^2 \leq x^2 \sin(\frac 1x) \leq x^2$

*and* $\lim_{x\to 0}-x^2 \leq \lim_{x\to 0}(x^2\sin(\frac 1x)) \leq \lim_{x\to 0}x^2$

$\lim_{x\to 0}-x^2 = 0$ *and* $\lim_{x\to 0}x^2 = 0$, so by Squeeze Theorem

$\lim_{x\to 0}(x^2\sin\frac 1x) = 0$

\(\lim_{\theta\to 0}\frac{1}{\cos\theta}\geq\lim_{\theta\to 0}\frac{\sin\theta}{\theta}\geq\lim_{\theta\to 0}\cos\theta\) \(1 \geq\ ? \geq 1\)

- Evaluate $\lim_{x\to 1} \arcsin(\frac{1-\sqrt x}{1-x})$
- Show analytically that $\lim_{x\to 0}x^2\cos(\frac{1}{x^2}) = 0$
- Evaluate $\lim_{x\to 0}\frac{(x+2)^4}{x}$
- Show analytically that $\lim_{x\to\infty}\frac{\sin x}{x} = 0$
- For the graphs below,
- Functions $f$, $g$, and $h$ are twice-differentiable functions with $g(2) = h(2) = 4$. The line $y = 4 + \frac 23 (x-2)$ is tangent to both the graph of $g$ at $x=2$ and the graph of $h$ at $x=2$.

It is known that $g(x) \leq h(x)$ for $1 < x < 3$. Let $k$ be a function satisfying $g(x)\leq k(x)\leq h(x)$ for $1 < x < 3$. Is $k$ continuous at $x=2$? Justify your answer.