User:JaviPrieto/Limits

!x!!$$\frac{\sin x}{x}$$
 * $$ \lim_{x \to p}f(x) = L, \, $$
 * $$ \lim_{x \to p}f(x) = L, \ $$
 * $$ \lim_{x \to p^+}f(x) = L $$
 * $$ \lim_{x \to p^-}f(x) = L $$
 * $$ \lim_{x \to p}f(x) = L $$
 * $$ \lim_{x \to p}f(x) = L $$
 * $$ \lim_{x \to p}f(x) = L $$
 * $$ \lim_{x \to \infty}f(x) = L,$$

if and only if for all $$\varepsilon > 0$$ there exists S > 0 such that $$|f(x) - L| < \varepsilon$$ whenever x > S.
 * $$ \lim_{x \to -\infty}f(x) = L,$$

if and only if for all $$\varepsilon > 0$$ there exists S < 0 such that $$|f(x) - L| < \varepsilon$$ whenever x < S.
 * $$ \lim_{x \to -\infty}e^x = 0. \, $$
 * $$ \lim_{x \to a}f(x) = \infty, \, $$

if and only if for all $$\varepsilon > 0$$ there exists $$\delta > 0$$ such that $$f(x) > \varepsilon$$ whenever $$|x - a| < \delta$$.
 * $$ \lim_{x \to \infty}f(x) = \infty, \lim_{x \to a^+}f(x) = -\infty. \, $$
 * $$\lim_{x \to 0^+}\ln x = -\infty. \, $$
 * $$\lim_{x \to 0^{+}}{1\over x} = \infty, \lim_{x \to \infty}{1\over x} = 0.$$

The complex plane with metric $$d(x, y) := |x-y|$$ is also a metric space. There are two different types of limits when the complex-valued functions are considered.
 * $$ \lim_{x \to p} f(x) = L $$

if and only if for all e > 0 there exists a d > 0 such that for all real numbers x with $$0 < |x - p| < \delta$$, then $$|f(x) - L| < \varepsilon$$.
 * $$ \lim_{(x,y) \to (p, q)} f(x, y) = L $$
 * $$\begin{matrix}\lim\limits_{x \to p} & (f(x) + g(x)) & = & \lim\limits_{x \to p} f(x) + \lim\limits_{x \to p} g(x) \\\lim\limits_{x \to p} & (f(x) - g(x)) & = & \lim\limits_{x \to p} f(x) - \lim\limits_{x \to p} g(x) \\\lim\limits_{x \to p} & (f(x)\cdot g(x)) & = & \lim\limits_{x \to p} f(x) \cdot \lim\limits_{x \to p} g(x) \\\lim\limits_{x \to p} & (f(x)/g(x)) & = & {\lim\limits_{x \to p} f(x) / \lim\limits_{x \to p} g(x)}\end{matrix}$$
 * $$\lim_{y \to d} f(y) = e$$, and $$\lim_{x \to c} g(x) = d \Rightarrow \lim_{x \to c} f(g(x)) = e$$,

is not true. However, this "chain rule" does hold if, in addition, either f(d) = e (i. e. f is continuous at d) or g does not take the value d near c (i. e. there exists a $$\delta >0$$ such that if $$0<|x-c|<\delta$$ then $$|g(x)-d|>0$$).
 * $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$
 * $$\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$$
 * $$\sin x < x < \tan x.$$
 * $$1 < \frac{x}{\sin x} < \frac{\tan x}{\sin x}$$
 * $$1 < \frac{x}{\sin x} < \frac{1}{\cos x}$$
 * $$\lim_{x \to 0} \frac{1}{\cos x} = \frac{1}{1} = 1$$
 * $$\lim_{x \to 0} \frac{x}{\sin x} = 1$$
 * $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

$$\lim_{x \to 0} \frac{\sin (2x)}{\sin (3x)} =\lim_{x \to 0} \frac{2 \cos (2x)}{3 \cos (3x)} =\frac{2 \sdot 1}{3 \sdot 1} =\frac{2}{3}.$$ A short way to write the limit $$\lim_{n \to \infty} \sum_{i=s}^{n} f(i) $$ is $$\sum_{i=s}^{\infty} f(i)$$. A short way to write the limit $$\lim_{x \to \infty} \int_{a}^{x} f(x) \; dx $$ is $$\int_{a}^{\infty} f(x) \; dx$$. A short way to write the limit $$\lim_{x \to -\infty} \int_{x}^{b} f(x) \; dx $$ is $$\int_{-\infty}^{b} f(x) \; dx$$.
 * $$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$