Dana jest funkcja:

$f\left(x\right)=x^2$  

a) Utwórzmy ciąg wartości (y_n) funkcji f dla ciągu argumentów (x_n), gdy:

$x_n=2-\frac{1}{n}$  

n	$x_n=2-\frac{1}{n}$	$y_n=f\left(x_n\right)={\left(x_n\right)}^2$
1	$x_n=2-1=1$	$y_1=f\left(x_1\right)=f\left(1\right)=1^2=1$
2	$x_n=2-\frac{1}{2}=\frac{3}{2}$	$y_2=f\left(x_2\right)=f\left(\frac{3}{2}\right)={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$
3	$x_n=2-\frac{1}{3}=\frac{5}{3}$	$y_3=f\left(x_3\right)=f\left(\frac{5}{3}\right)={\left(\frac{5}{3}\right)}^2=\frac{25}{9}$
4	$x_n=2-\frac{1}{4}=\frac{7}{4}$	$y_4=f\left(x_4\right)=f\left(\frac{7}{4}\right)={\left(\frac{7}{4}\right)}^2=\frac{49}{16}$
...	$\dots$	$\dots$
n	$x_n=2-\frac{1}{n}$	$y_n=f\left(x_n\right)=f\left(2-\frac{1}{n}\right)={\left(2-\frac{1}{n}\right)}^2$

Stąd:

$\left(x_n\right)=\left(1,\frac{3}{2},\frac{5}{3},\frac{7}{4},\dots ,2-\frac{1}{n}\right)$  

$\left(y_n\right)=\left(1,\frac{9}{4},\frac{25}{9},\frac{49}{16},\dots ,{\left(2-\frac{1}{n}\right)}^2\right)$  

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b) Utwórzmy ciąg wartości (y_n) funkcji f dla ciągu argumentów (x_n), gdy:

$x_n=2+\frac{{\left(-1\right)}^n}{n}$  

n	$x_n=2+\frac{{\left(-1\right)}^n}{n}$	$y_n=f\left(x_n\right)={\left(x_n\right)}^2$
1	$x_n=2-1=1$	$y_1=f\left(x_1\right)=f\left(1\right)=1^2=1$
2	$x_n=2+\frac{1}{2}=\frac{5}{2}$	$y_2=f\left(x_2\right)=f\left(\frac{5}{2}\right)={\left(\frac{5}{2}\right)}^2=\frac{25}{4}$
3	$x_n=2-\frac{1}{3}=\frac{5}{3}$	$y_3=f\left(x_3\right)=f\left(\frac{5}{3}\right)={\left(\frac{5}{3}\right)}^2=\frac{25}{9}$
4	$x_n=2+\frac{1}{4}=\frac{9}{4}$	$y_4=f\left(x_4\right)=f\left(\frac{9}{4}\right)={\left(\frac{9}{4}\right)}^2=\frac{81}{16}$
...	$\dots$	$\dots$
n	$x_n=2+\frac{{\left(-1\right)}^n}{n}$	$y_n=f\left(x_n\right)=f\left(2+\frac{{\left(-1\right)}^n}{n}\right)={\left(2+\frac{{\left(-1\right)}^n}{n}\right)}^2$

Stąd:

$\left(x_n\right)=\left(1,\frac{5}{2},\frac{5}{3},\frac{9}{4},\dots ,2+\frac{{\left(-1\right)}^n}{n}\right)$  

$\left(y_n\right)=\left(1,\frac{25}{4},\frac{25}{9},\frac{81}{16},\dots ,{\left(2+\frac{{\left(-1\right)}^n}{n}\right)}^2\right)$