Granice funkcji

[aleksandrapyrpec]

Proszę o pomoc bo jutro kolokwium

\(\lim_{x\to -1} \frac{x^n - 1}{x-1}\)

\(\lim_{x\to 25} \frac{ \sqrt{x} -5}{x - 25}\)

\(\lim_{x\to 0} \frac{ \sqrt{x^2 + 1} - \sqrt{x + 1} }{1 - \sqrt{x + 1} }\)

\(\lim_{x\to 0} \frac{4x}{3sin2x}\)

\(\lim_{x\to 8} \frac{8 - x}{sin (\frac{1}{8}) \pi x }\)

[radagast]

\(\lim_{x\to -1} \frac{x^n - 1}{x-1}=\lim_{x\to -1} \frac{(x- 1)(x^{n-1}+x^{n-2}+....+1)}{x-1}=\lim_{x\to -1} (x^{n-1}+x^{n-2}+....+1)= \begin{cases} 1\ \ \ \ dla\ n\ nieparzystych\\0 \ \ \ \ dla\ n\ parzystych\end{cases}\)

[radagast]

\(\lim_{x\to 25} \frac{ \sqrt{x} -5}{x - 25}=\lim_{x\to 25} \frac{ (\sqrt{x} -5)(\sqrt{x} +5)}{(x - 25)(\sqrt{x} +5)}=\lim_{x\to 25} \frac{1}{\sqrt{x} +5}= \frac{1}{10}\)

[radagast]

\(\lim_{x\to 0} \frac{ \sqrt{x^2 + 1} - \sqrt{x + 1} }{1 - \sqrt{x + 1} }=\lim_{x\to 0} \frac{ (\sqrt{x^2 + 1} - \sqrt{x + 1}) (\sqrt{x^2 + 1} + \sqrt{x + 1}) (1 + \sqrt{x + 1})}{(1 - \sqrt{x + 1}) (1 + \sqrt{x + 1})(\sqrt{x^2 + 1} + \sqrt{x + 1}) }=
\lim_{x\to 0} \frac{(x^2-x) (1 + \sqrt{x + 1})}{x(\sqrt{x^2 + 1} + \sqrt{x + 1}) }=\lim_{x\to 0} \frac{(x-1) (1 + \sqrt{x + 1})}{(\sqrt{x^2 + 1} + \sqrt{x + 1}) }= \frac{-1 \cdot 2}{1+1}=-1\)

[radagast]

\(\lim_{x\to 0} \frac{4x}{3sin2x}=\lim_{x\to 0} \frac{2 \cdot 2x}{3sin2x}= \frac{2}{3}\lim_{x\to 0} \frac{2x}{sin2x}= \frac{2}{3} \cdot 1= \frac{2}{3}\)

[radagast]

\(\lim_{x\to 8} \frac{8 - x}{sin (\frac{\pi x}{8}) }= \left(8-x=t\\x=t-8\\ \lim_{x\to 8} t=0 \right)=\lim_{t\to 0} \frac{t}{sin (\frac{\pi (t-8)}{8}) }=\lim_{t\to 0} \frac{t}{sin (\frac{\pi t}{8}- \pi ) }=\lim_{t\to 0} \frac{t}{-sin {\frac{\pi t}{8}} }= \left( \right) \left( \frac{\pi t}{8} =z\\t= \frac{8z}{ \pi } \right)=
\frac{8}{ \pi }\lim_{z\to 0} \frac{ z}{-sin {z} }=- \frac{8}{ \pi }\)
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