Esercizio
$\lim_{n\to\infty}\left(\frac{2n+1}{1-3\sqrt{n}}\right)$
Soluzione passo-passo
Impara online a risolvere i problemi di integrali trigonometrici passo dopo passo. (n)->(infinito)lim((2n+1)/(1-3n^(1/2))). Applicare la formula: \lim_{x\to c}\left(\frac{a}{b}\right)=\lim_{x\to c}\left(\frac{\frac{a}{sign\left(c\right)fgrow\left(b\right)}}{\frac{b}{sign\left(c\right)fgrow\left(b\right)}}\right), dove a=2n+1, b=1-3\sqrt{n}, c=\infty , a/b=\frac{2n+1}{1-3\sqrt{n}}, x=n e x->c=n\to\infty . Applicare la formula: \lim_{x\to c}\left(\frac{a}{b}\right)=\lim_{x\to c}\left(\frac{radicalfrac\left(a\right)}{radicalfrac\left(b\right)}\right), dove a=\frac{2n+1}{\sqrt{n}}, b=\frac{1-3\sqrt{n}}{\sqrt{n}}, c=\infty e x=n. Applicare la formula: \lim_{x\to c}\left(\frac{a}{b}\right)=\lim_{x\to c}\left(\frac{splitfrac\left(a\right)}{splitfrac\left(b\right)}\right), dove a=\sqrt{\frac{n}{\left(2n+1\right)^{2}}}, b=\sqrt{\frac{n}{\left(1-3\sqrt{n}\right)^{2}}}, c=\infty e x=n. Applicare la formula: \frac{\frac{a}{b}}{\frac{c}{f}}=\frac{af}{bc}, dove a=n, b=\left(2n+1\right)^{2}, a/b/c/f=\frac{\frac{n}{\left(2n+1\right)^{2}}}{\frac{n}{\left(1-3\sqrt{n}\right)^{2}}}, c=n, a/b=\frac{n}{\left(2n+1\right)^{2}}, f=\left(1-3\sqrt{n}\right)^{2} e c/f=\frac{n}{\left(1-3\sqrt{n}\right)^{2}}.
(n)->(infinito)lim((2n+1)/(1-3n^(1/2)))
Risposta finale al problema
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