Mathematics Crystal - mat/Topics/Differentiation/Limits.tex

/ Topics / Differentiation / Limits.tex PDF

\documentclass[10pt]{article}
\usepackage{amssymb,amsmath}
\usepackage[hmargin=1cm,vmargin=1cm]{geometry}
\begin{document}
{\large Limits}
\begin{align*}
\text{In the following,}\quad&\lim_{x\to a}f(x)=F\quad\text{and}\quad\lim_{x\to a}g(x)=G\:.\\
\\
&\lim_{x\to a}kf(x)=k\times\lim_{x\to a}f(x)=kF\\
&\lim_{x\to a}[f(x)\pm g(x)]=\lim_{x\to a}f(x)\pm\lim_{x\to a}g(x)=F\pm G\\
&\lim_{x\to a}[f(x)\cdot g(x)]=\lim_{x\to a}f(x)\times\lim_{x\to a}g(x)=F\cdot G\\
&\lim_{x\to a}[f(x)]^n=[F]^n\\
&\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}g(x)}=\frac{F}{G}\:,\quad\text{where }G\neq 0\:.\\
&\lim_{x\to a}\left[e^{f(x)}\right]=e^{\:\lim\limits_{x\to a}f(x)}=e^F\\
&\lim_{x\to a}[\log_b f(x)]=\log_b\left[\lim_{x\to a}f(x)\right]=\log_b F\:,\quad\text{where }F>0\:.\\
&\lim_{x\to\pm\infty}\frac{\:1\:}{x}=0\\
&\lim_{x\to\pm\infty}\frac{\:1\:}{x^n}=0\:,\quad\text{where }n>0\:.\\
&\lim_{x\to 0^+}\frac{\:1\:}{x}=+\infty\\
&\lim_{x\to 0^-}\frac{\:1\:}{x}=-\infty\\
\\
&\sin x<x<\tan x\:,\quad\text{where }0<x<\frac{\:\pi\:}{2}\\
&\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{x}{\sin x}=1\\
&\lim_{x\to 0}\frac{\tan x}{x}=\lim_{x\to 0}\frac{x}{\tan x}=1\\
\\
&\lim_{x\to 0}\frac{1-\cos x}{x}=0\qquad
\left(\tfrac{1-\cos x}{x}=\tfrac{1-\cos x}{x}\cdot\tfrac{1+\cos x}{1+\cos x}=\tfrac{\sin^2 x}{x(1+\cos x)}=\tfrac{\sin x}{x}\tfrac{\sin x}{1+\cos x}\right)\\
&\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{\:1\:}{2}\qquad
\left(\tfrac{1-\cos x}{x^2}=\tfrac{1-\cos x}{x^2}\cdot\tfrac{1+\cos x}{1+\cos x}=\tfrac{\sin^2 x}{x^2(1+\cos x)}=\tfrac{\sin^2 x}{x^2}\tfrac{1}{1+\cos x}\right)\\
\end{align*}
\end{document}