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August 2022 1 23 Report

Вычислите приделы, пожалуйста если можите распешите на листочке
[tex]1) \lim_{n \to \ 0} \frac{1-\sqrt{1-x^{2} } }{x^{2} } \\2) \lim_{n \to \infty} \frac{5x^{4}-x^{3}+2x}{x^{4}-8x^{3}+1} \\3) \lim_{n \to \infty} (1+\frac{3}{x} )^{-x}[/tex]

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NNNLLL54

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$1)\ \ \lim\limits _{x \to 0}\dfrac{1-\sqrt{1-x^2}}{x^2}=\lim\limits _{x \to 0}\dfrac{(1-\sqrt{1-x^2})(1+\sqrt{1-x^2})}{x^2\, (1+\sqrt{1-x^2})}=\\\\\\=\lim\limits _{x \to 0}\, \dfrac{1-(1-x^2)}{x^2\, (1+\sqrt{1-x^2})}=\lim\limits _{x \to 0}\dfrac{x^2}{x^2\, (1+\sqrt{1-x^2})}=\lim\limits _{x \to 0}\dfrac{1}{1+\sqrt{1+x^2}}=\dfrac{1}{2}$

$2)\ \ \lim\limits _{x \to \infty}\dfrac{5x^4-x^3+2x}{x^4-8x^3+1}=\lim\limits _{x \to \infty}\, \dfrac{5x^4}{1\cdot x^4}=\dfrac{5}{1}=5\\\\\\3)\ \ \lim\limits _{x \to \infty}\Big(1+\dfrac{3}{x}\Big)^{-x}=\lim\limits _{x \to \infty}\Big(\underbrace{\Big(1+\dfrac{3}{x}\Big)^{\frac{x}{3}}}_{\to \, e}\Big)\, ^{-3}=e^{-3}=\dfrac{1}{e^3}$

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