Функције – граничне вредности функција 4

Текст задатака објашњених у видео лекцији.

Одредити граничне вредности:

пр.12)   $\mathop {\lim }\limits_{x \to \infty } (x - \sqrt {{x^2} - 8x} )$

пр.13)   $\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} - 4x + 5}  - \sqrt {{x^2} + 4x + 5} )$

пр.12)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } (x - \sqrt {{x^2} - 8x} ) = \mathop {\lim }\limits_{x \to \infty } (x - \sqrt {{x^2} - 8x} ) \cdot \frac{{x + \sqrt {{x^2} - 8x} }}{{x + \sqrt {{x^2} - 8x} }} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {x - \sqrt {{x^2} - 8x} } \right)\left( {x + \sqrt {{x^2} - 8x} } \right)}}{{x + \sqrt {{x^2} - 8x} }} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - {{\left( {\sqrt {{x^2} - 8x} } \right)}^2}}}{{x + \sqrt {{x^2} - 8x} }} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - {x^2} + 8x}}{{x + \sqrt {{x^2} - 8x} }} = \mathop {\lim }\limits_{x \to \infty } \frac{{8x}}{{x + \sqrt {{x^2} - 8x} }} \cdot \frac{x}{x} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{8}{{1 + \sqrt {1 - \frac{{8x}}{{{x^2}}}} }} = \frac{8}{{1 + \sqrt 1 }} = 4 \hfill \\
\end{gathered} \]

пр.13)  

\[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} - 4x + 5} - \sqrt {{x^2} + 4x + 5} ) = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} - 4x + 5} - \sqrt {{x^2} + 4x + 5} ) \cdot \frac{{\sqrt {{x^2} - 4x + 5} + \sqrt {{x^2} + 4x + 5} }}{{\sqrt {{x^2} - 4x + 5} + \sqrt {{x^2} + 4x + 5} }} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - 4x + 5 - \left( {{x^2} + 4x + 5} \right)}}{{\sqrt {{x^2} - 4x + 5} + \sqrt {{x^2} + 4x + 5} }} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - 4x + 5 - {x^2} - 4x - 5}}{{\sqrt {{x^2} - 4x + 5} + \sqrt {{x^2} + 4x + 5} }} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{ - 8x}}{{\sqrt {{x^2} - 4x + 5} + \sqrt {{x^2} + 4x + 5} }} \cdot \frac{x}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - \frac{{8x}}{x}}}{{\sqrt {\frac{{{x^2} - 4x + 5}}{{{x^2}}}} + \sqrt {\frac{{{x^2} + 4x + 5}}{{{x^2}}}} }} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{ - 8}}{{\sqrt {1 - \frac{4}{x} + \frac{5}{{{x^2}}}} + \sqrt {1 + \frac{4}{x} + \frac{5}{{{x^2}}}} }} = \frac{{ - 8}}{2} = - 4 \hfill \\
\end{gathered} \]