Bardzo bym prosił o rozwiązanie któryś z tych podpunktów kto rozwiąże najwiecej dam maksa:) Pozdrawi.... Question from @Kravis

September 2018 1 19 Report

Bardzo bym prosił o rozwiązanie któryś z tych podpunktów kto rozwiąże najwiecej dam maksa:) Pozdrawiam :*

zadanie 1

$\lim_{x \rightarrow -1} \frac{x^3 + x^2+3x+3}{x^2-2x-3} = \lim_{x \rightarrow -1} \frac{x^2(x + 1) + 3(x + 1)}{(x + 1)(x - 3)} = \lim_{x \rightarrow -1} \frac{(x^2 + 3)(x + 1)}{(x + 1)(x - 3)} = \lim_{x \rightarrow -1} \frac{x^2 + 3}{x - 3} = \frac{1 + 3}{-1-3} = - 1$

$\lim_{x \rightarrow 1} \frac{x^5 - 1}{x - 1} = \lim_{x \rightarrow 1} \frac{(x - 1)(x^4 + x^3 + x^2 + 1)}{x - 1} = \lim_{x \rightarrow 1} (x^4 + x^3 + x^2 + 1) = 1 + 1 + 1 + 1 = 4$

$\lim_{x \rightarrow 4} \frac{x - 4}{2 - \sqrt{x}} = \lim_{x \rightarrow 4} \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{2 - \sqrt{x}} = \lim_{x \rightarrow 4} -(\sqrt{x} + 2) = -(\sqrt{4} + 2) = - 4$

$\lim_{x \rightarrow 0} \frac{\sqrt{x + 1} - 1}{x} = \lim_{x \rightarrow 0} \frac{\left(\sqrt{x+1} - 1\right)\left(\sqrt{x+1}+1\right)}{x\left(\sqrt{x+1}+1\right)} = \lim_{x \rightarrow 0} \frac{x + 1 - 1}{x\left(\sqrt{x+1}+1\right)} = \lim_{x \rightarrow 0} \frac{x}{x\left(\sqrt{x+1}+1\right)} = \lim_{x \rightarrow 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{\sqrt{0+1}+1} = \frac{1}{2}$

$\lim_{x \rightarrow \infty} \left(\frac{x^3+x^2}{x^2+4} - x\right) = \lim_{x \rightarrow \infty} x\left(\frac{x^2 + x}{x^2+4} - 1\right) = \lim_{x \rightarrow \infty} x\left(\frac{x^2 +4 - 4 + x}{x^2+4} - 1\right) = \lim_{x \rightarrow \infty} x\left(1 + \frac{x - 4}{x^2+4} - 1\right) = \lim_{x \rightarrow \infty} x*\frac{x - 4}{x^2+4} = \lim_{x \rightarrow \infty} \frac{x^2\left(1 - \frac{4}{x^2}\right)}{x^2\left(1+\frac{4}{x^2}\right)} = \lim_{x \rightarrow \infty} \frac{1 - \frac{4}{x^2}}{1+\frac{4}{x^2}} = 1$

zadanie 2

dziedzina - asymaptota pionowa:

$x^2 \neq 0\\ x \neq 0\\ lim_{x \rightarrow 0^+} \frac{2x^3 - 1}{x^2} = \left[\frac{-1}{0^+}\right] = -\infty\\ lim_{x \rightarrow 0^-} \frac{2x^3 - 1}{x^2} = \left[\frac{-1}{0^+}\right] = -\infty$

mamy obustronna asymptotę pionową

asymptota ukośna: y = ax + b

$a = lim_{x \rightarrow +\infty}\frac{f(x)}{x} = lim_{x \rightarrow +\infty} \frac{2x^3 - 1}{x^3} = lim_{x \rightarrow +\infty} \frac{2 - \frac{1}{x^3}}{1} = 2\\ b = lim_{x \rightarrow +\infty}(f(x) - ax) = lim_{x \rightarrow +\infty} \left(\frac{2x^3 - 1}{x^2} - 2x\right) = lim_{x \rightarrow +\infty} \frac{2x^3 - 1 - 2x^3}{x^2} = lim_{x \rightarrow +\infty} -\frac{1}{x^2} = 0\\ a = lim_{x \rightarrow -\infty}\frac{f(x)}{x} = lim_{x \rightarrow -\infty} \frac{2x^3 - 1}{x^3} = lim_{x \rightarrow -\infty} \frac{2 - \frac{1}{x^3}}{1} = 2\\ b = lim_{x \rightarrow -\infty}(f(x) - ax) = lim_{x \rightarrow -\infty} \left(\frac{2x^3 - 1}{x^2} - 2x\right) = lim_{x \rightarrow -\infty} \frac{2x^3 - 1 - 2x^3}{x^2} = lim_{x \rightarrow -\infty} -\frac{1}{x^2} = 0\\ y = 2x$

jedna obustronna asymptota ukośna (jest ukośna bo a ≠ 0, więc brak poziomej)

dziedzina - asymaptota pionowa:

$\sqrt{x^2 - 1} \neq 0\\ x \neq 0 \wedge x \neq - 1\\ \\ x^2 - 1 \geq 0\\ x \in \left(-\infty, - 1\right> \cup \left<1, \infty\right)\\ \\ D: \ x \in \left(-\infty, - 1\right) \cup \left(1, \infty\right)\\ \\ lim_{x \rightarrow - 1^-} \frac{x^2}{\sqrt{x^2 - 1}} = \left[\frac{1}{0^+}\right] =+ \infty\\ lim_{x \rightarrow 1^+} \frac{x^2}{\sqrt{x^2 - 1}} = \left[\frac{1}{0^+}\right] =+ \infty$

\mamy dwie jednostronne asymptoty:

x = - 1 - lewostronna

x = 1 - prawostronna

asymptota ukośna: y = ax + b

$a = \lim_{x\rightarrow -\infty}\frac{f(x)}{x} = \lim_{x\rightarrow -\infty} \frac{x}{\sqrt{x^2 - 1}} = \lim_{x\rightarrow -\infty} \frac{x}{x\sqrt{1 - \frac{1}{x^2}}} = a = \lim_{x\rightarrow -\infty}\frac{f(x)}{x} = \lim_{x\rightarrow -\infty} \frac{x}{\sqrt{x^2 - 1}} = \lim_{x\rightarrow -\infty} \frac{1}{\sqrt{1 - \frac{1}{x^2}}} = \frac{1}{\sqrt{1 - 0}} = 1\\ b = \lim_{x\rightarrow -\infty}(f(x) - ax) = \lim_{x\rightarrow -\infty} \left(\frac{x^2}{\sqrt{x^2 - 1}} - x\right) = \lim_{x\rightarrow -\infty} \left(\frac{x^2\sqrt{x^2 - 1}}{x^2 - 1} - x\right) = \lim_{x\rightarrow -\infty} \frac{x^2\sqrt{x^2 - 1} - x^2 + 1}{x^2 - 1} = \lim_{x\rightarrow -\infty} \frac{x^2\left(\sqrt{x^2 - 1} - 1 + \frac{1}{x^2}\right)}{x^2\left(1 - \frac{1}{x^2}\right)} = \infty\\ a = \lim_{x\rightarrow +\infty}\frac{f(x)}{x} = \lim_{x\rightarrow +\infty} \frac{x}{\sqrt{x^2 - 1}} = \lim_{x\rightarrow +\infty} \frac{x}{x\sqrt{1 - \frac{1}{x^2}}} = a = \lim_{x\rightarrow +\infty}\frac{f(x)}{x} = \lim_{x\rightarrow +\infty} \frac{x}{\sqrt{x^2 - 1}} = \lim_{x\rightarrow +\infty} \frac{1}{\sqrt{1 - \frac{1}{x^2}}} = \frac{1}{\sqrt{1 - 0}} = 1\\ b = \lim_{x\rightarrow +\infty}(f(x) - ax) = \lim_{x\rightarrow +\infty} \left(\frac{x^2}{\sqrt{x^2 - 1}} - x\right) = \lim_{x\rightarrow +\infty} \left(\frac{x^2\sqrt{x^2 - 1}}{x^2 - 1} - x\right) = \lim_{x\rightarrow +\infty} \frac{x^2\sqrt{x^2 - 1} - x^2 + 1}{x^2 - 1} = \lim_{x\rightarrow +\infty} \frac{x^2\left(\sqrt{x^2 - 1} - 1 + \frac{1}{x^2}\right)}{x^2\left(1 - \frac{1}{x^2}\right)} = \infty$

brak asymptoty poziomej/ ukośnej

zadanie 3

czyli czy p jest granicą w odpowiednim punkcie:

$\lim_{x\rightarrow 1} \frac{\sqrt{x} - 1}{x - 1} = \lim_{x\rightarrow 1} \frac{\sqrt{x} - 1}{(\sqrt{x} - 1)(\sqrt{x} + 1)} = \lim_{x\rightarrow 1} \frac{1}{\sqrt{x}+1} = \frac{1}{2} = p$

$\lim_{x\rightarrow 0} \frac{\sqrt{2x+1}-\sqrt{x+1}}{x} = \lim_{x\rightarrow 0} \frac{\sqrt{2x+1}-\sqrt{x+1}}{x}*\frac{\sqrt{2x+1}+\sqrt{x+1}}{\sqrt{2x+1}+\sqrt{x+1}} = \lim_{x\rightarrow 0} \frac{2x + 1 - x - 1}{x\left(\sqrt{2x+1}+\sqrt{x+1}\right)} = \lim_{x\rightarrow 0} \frac{x}{x\left(\sqrt{2x+1}+\sqrt{x+1}\right)} = \lim_{x\rightarrow 0} \frac{1}{\sqrt{2x+1}+\sqrt{x+1}} =\frac{1}{2} = p$