Question

Oblicz:

a) [tex]\lim_{n \to \infty} (\sqrt{n^{2}-1 }-\sqrt{2n^{2}-1})[/tex]

b) [tex]\lim_{n \to \infty} (3-2n^2-n^3)[/tex]

c) [tex]\lim_{n \to \infty} \frac{n^2-2}{n+1}[/tex]

Answer 1

[tex]\huge\begin{array}{ccc}a)&\lim\limits_{n\to\infty} a_n=-\infty\\b)&\lim\limits_{n\to\infty} b_n=-\infty\\c)&\lim\limits_{n\to\infty} c_n=\infty\end{array}[/tex]

Granica ciągu.

Wiemy, że:

[tex]\dfrac{a}{n}\xrightarrow{n\to\infty}0[/tex]

ROZWIĄZANIE:

a)

[tex]\lim\limits_{n\to\infty}\left(\sqrt{n^2-1}-\sqrt{2n^2-1}\right)=\lim\limits_{n\to\infty}\left(\sqrt{n^2\left(1-\frac{1}{n^2}\right)}-\sqrt{n^2\left(2-\frac{1}{n}\right)}\right)\\\\=\lim\limits_{n\to\infty}\left(\sqrt{n^2}\cdot\sqrt{1-\frac{1}{n^2}}-\sqrt{n^2}\cdot\sqrt{2-\frac{1}{n^2}}\right)\\\\=\lim\limits_{n\to\infty}\left(n\sqrt{1-\frac{1}{n^2}}-n\sqrt{2-\frac{1}{n^2}}\right)[/tex]

[tex]=\lim\limits_{n\to\infty}n\left(\sqrt{1-\frac{1}{n^2}}-\sqrt{2-\frac{1}{n^2}}\right)[/tex]

[tex]\dfrac{1}{n^2}\xrightarrow{n\to\infty}0\\\\\Downarrow\\\\\sqrt{1-\frac{1}{n^2}}\xrightarrow{n\to\infty}\sqrt{1-0}=\sqrt1=1\ \wedge\sqrt{2-\frac{1}{n^2}}\xrightarrow{n\to\infty}\sqrt{2-0}=\sqrt2[/tex]

Jako, że:

[tex]1 < \sqrt2\to1-\sqrt2 < 0[/tex]

W związku z tym:

[tex]\lim\limits_{n\to\infty}\left(\sqrt{n^2-1}-\sqrt{2n^2-1}\right)=-\infty[/tex]

b)

[tex]\lim\limits_{n\to\infty}\left(3-2n^2-n^3\right)=\lim\limits_{n\to\infty}n^3\left(\dfrac{3}{n^3}-\dfrac{2}{n}-1\right)[/tex]

[tex]n^3\xrightarrow{n\to\infty}\infty\\\\\dfrac{3}{n^3}\xrightarrow{n\to\infty}0\\\\-\dfrac{2}{n^2}\xrightarrow{n\to\infty}0\\\\-1\xrightarrow{n\to\infty}-1[/tex]

W związku z tym:

[tex]\lim\limits_{n\to\infty}\left(3-2n^2-n^3\right)=-\infty[/tex]

c)

[tex]\lim\limits_{n\to\infty}\dfrac{n^2-2}{n+1}=\lim\limits_{n\to\infty}\dfrac{n(n-2)}{n\left(1+\frac{1}{n}\right)}=\lim\limits_{n\to\infty}\dfrac{n-2}{1+\frac{1}{n}}[/tex]

[tex]\dfrac{1}{n}\xrightarrow{n\to\infty}0\\\\1-\dfrac{1}{n}\xrightarrow{n\to\infty}1-0=1\\\\n-2\xrightarrow{n\to\infty}\infty[/tex]

W związku z tym:

[tex]\lim\limits_{n\to\infty}\dfrac{n^2-2}{n+1}=\infty[/tex]