Racionální lomené funkce
Úloha číslo: 2954
Určete limity následujících funkcí ve všech bodech mimo definiční obor a v \(\pm\infty\):
Varianta 1
\(\displaystyle \frac{3x^2+1}{2x^2+1} \)
Řešení
\(D_f=\mathbb R\)
\(\displaystyle \lim_{x\to\pm\infty}\frac{3x^2+1}{2x^2+1}= \frac{3+\lim_{x\to\pm\infty}x^{-2}}{2+\lim_{x\to\pm\infty}x^{-2}}= \frac{3+0}{2+0}= \frac32 \)
Varianta 2
\(\displaystyle \frac{2x^2+1}{x^3+1} \)
Řešení
\(\displaystyle \frac{2x^2+1}{x^3+1}= \frac{2x^2+1}{(x+1)(x^2-x+1)} \), tedy \(D_f=\mathbb R\setminus\{-1\}\)
\(\displaystyle \lim_{x\to -1^-}\frac{2x^2+1}{x^3+1}= \lim_{x\to -1^-}\frac{2x^2+1}{(x+1)(x^2-x+1)}= \frac31\lim_{x\to-1^-}\frac{1}{x+1}= -\infty \)
\(\displaystyle \lim_{x\to-1^+}\frac{2x^2+1}{x^3+1}= \lim_{x\to-1^+}\frac{2x^2+1}{(x+1)(x^2-x+1)}= \frac31\lim_{x\to-1^+}\frac{1}{x+1}= \infty \)
\(\displaystyle \lim_{x\to\pm\infty}\frac{2x^2+1}{x^3+1}= \frac{\lim_{x\to\pm\infty}(2x^{-1}+x^{-3})}{1+\lim_{x\to\pm\infty}x^{-3}}= \frac{0}{1+0}=0 \)
Varianta 3
\(\displaystyle \frac{x^2-1}{2x^2-x-1} \)
Řešení
\(\displaystyle \frac{x^2-1}{2x^2-x-1}= \frac{(x-1)(x+1)}{(x-1)(2x+1)} \), tedy \(D_f=\mathbb R\setminus\{-\frac12{,}1\}\)
\(\displaystyle \lim_{x\to1}\frac{x^2-1}{2x^2-x-1}= \lim_{x\to1}\frac{(x-1)(x+1)}{(x-1)(2x+1)}= \lim_{x\to1}\frac{x+1}{2x+1}= \frac23 \)
\(\displaystyle \lim_{x\to-\frac12^-}\frac{x^2-1}{2x^2-x-1}= \lim_{x\to-\frac12^-}\frac{x^2-1}{(x-1)(2x+1)}= \frac12\lim_{x\to-\frac12^-}\frac{1}{2x+1}= \frac12\cdot(-\infty)= -\infty \)
\(\displaystyle \lim_{x\to-\frac12^+}\frac{x^2-1}{2x^2-x-1}= \lim_{x\to-\frac12^+}\frac{x^2-1}{(x-1)(2x+1)}= \frac12\lim_{x\to-\frac12^+}\frac{1}{2x+1}= \frac12\cdot\infty= \infty \)
\(\displaystyle \lim_{x\to\pm\infty}\frac{x^2-1}{2x^2-x-1}= \frac{1-\lim_{x\to\pm\infty}x^{-2}}{2-\lim_{x\to\pm\infty}(x^{-1}+x^{-2})}= \frac{1-0}{2-0}=\frac12 \)
Varianta 4
\(\displaystyle \frac{2x^2+1}{x^2-1} \)
Řešení
\(\displaystyle \frac{2x^2+1}{x^2-1}= \frac{2x^2+1}{(x-1)(x+1)} \), tedy \(D_f=\mathbb R\setminus\{-1{,}1\}\)
\(\displaystyle \lim_{x\to-1^-}\frac{2x^2+1}{x^2-1}= \lim_{x\to-1^-}\frac{2x^2+1}{(x-1)(x+1)}= \frac3{-2}\lim_{x\to-1^-}\frac{1}{x+1}= \frac3{-2}\cdot(-\infty)= \infty \)
\(\displaystyle \lim_{x\to-1^+}\frac{2x^2+1}{x^2-1}= \lim_{x\to-1^+}\frac{2x^2+1}{(x-1)(x+1)}= \frac3{-2}\lim_{x\to-1^+}\frac{1}{x+1}= \frac3{-2}\cdot\infty= -\infty \)
\(\displaystyle \lim_{x\to1^-}\frac{2x^2+1}{x^2-1}= \lim_{x\to1^-}\frac{2x^2+1}{(x-1)(x+1)}= \frac3{2}\lim_{x\to1^-}\frac{1}{x-1}= \frac3{2}\cdot(-\infty)= -\infty \)
\(\displaystyle \lim_{x\to1^+}\frac{2x^2+1}{x^2-1}= \lim_{x\to1^+}\frac{2x^2+1}{(x-1)(x+1)}= \frac3{2}\lim_{x\to1^+}\frac{1}{x-1}= \frac3{2}\cdot\infty= \infty \)
\(\displaystyle \lim_{x\to\pm\infty}\frac{2x^2+1}{x^2-1}= \frac{2+\lim_{x\to\pm\infty}x^{-2}}{1-\lim_{x\to\pm\infty}x^{-2}}= \frac{2+0}{1+0}=2 \)
