Kilka zadań z ciągów

[Adrian47]

Witam. Mógłby mi objaśnić ktoś jak obliczyć te granice? Bo Wolfram głupieje

• \( a_n = \sqrt{(n^2 + 5n + 1)} - \sqrt{(5n^2 - n)}\)

• \( a_n =(\frac{3n-7}{3n+5})^{4n+3}\)

• \( \Lim_{x\to 0} f(x)=\frac{sin4x}{5x^2 + 3x}\)

[eresh]

Witam. Mógłby mi objaśnić ktoś jak obliczyć te granice? Bo Wolfram głupieje

• \( a_n = \sqrt{(n^2 + 5n + 1)} - \sqrt{(5n^2 - n)}\)

\(a_n = \sqrt{(n^2 + 5n + 1)} - \sqrt{(5n^2 - n)}=\frac{\sqrt{(n^2 + 5n + 1)} - \sqrt{(5n^2 - n)})(\sqrt{(n^2 + 5n + 1)} + \sqrt{(5n^2 - n)})}{\sqrt{(n^2 + 5n + 1)} + \sqrt{(5n^2 - n)}}=\\=\frac{n^2+5n+1-5n^2+n}{\sqrt{(n^2 + 5n + 1)} + \sqrt{(5n^2 - n)}}=\frac{-4n^n+6n+1}{\sqrt{(n^2 + 5n + 1)} + \sqrt{(5n^2 - n)}}=\frac{n^2(-4+\frac{6}{n^2}+\frac{1}{n^2})}{n(\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{5-\frac{1}{n}})}=\\=\frac{n(-4+\frac{6}{n^2}+\frac{1}{n^2})}{(\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{5-\frac{1}{n}})}\to [\frac{\infty\cdot (-4)}{\sqrt{1}+\sqrt{5}}]=-\infty\)

[radagast]

• \( \Lim_{x\to 0} f(x)=\frac{sin4x}{5x^2 + 3x}\)

\( \Lim_{x\to 0} f(x)=\Lim_{x\to 0}\frac{\sin 4x}{5x^2 + 3x}=\Lim_{x\to 0}\frac{\sin 4x}{4x} \frac{4x}{5x^2 + 3x} =\Lim_{x\to 0}\frac{\sin 4x}{4x} \frac{4}{5x + 3} = \frac{4}{3} \)

[radagast]

• \( a_n =(\frac{3n-7}{3n+5})^{4n+3}\)

\( \Lim_{x\to \infty } (\frac{3n-7}{3n+5})^{4n+3}= \Lim_{x\to \infty } (\frac{3n+5-12}{3n+5})^{4n+3}=\Lim_{x\to \infty } (1-\frac{12}{3n+5})^{(3n+5) \frac{4n+3}{3n+5} }=e^{-12 \cdot \frac{4}{3}}=e^{-16} \)