пределы нужно решить. Created by mILEENNSSSSSS7. algebra-ru

пределы нужно решить...

July 2022 1 2 Report

пределы нужно решить

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nikebod313

$6)\ \lim_{x \to 0} \dfrac{\sqrt{x + 25} - 5 }{x^{2} + 2x} = \bigg \{ \dfrac{0}{0} \bigg\} = \lim_{x \to 0} \dfrac{(\sqrt{x + 25} - 5)(\sqrt{x + 25} + 5) }{(x^{2} + 2x)(\sqrt{x + 25} + 5)} =\\\\= \lim_{x \to 0} \dfrac{(\sqrt{x + 25} )^{2} - 5^{2}}{(x^{2} + 2x)(\sqrt{x + 25} + 5)} = \lim_{x \to 0} \dfrac{x}{x(x + 2)(\sqrt{x + 25} + 5)} =\\\\= \lim_{x \to 0} \dfrac{1}{(x + 2)(\sqrt{x + 25} + 5)} = \dfrac{1}{(0+2)(\sqrt{0 + 25} + 5)} = \dfrac{1}{20}$

$7) \ \lim_{x \to \infty} \dfrac{x + 5x^{2} - x^{3}}{2x^{3} - x^{2} + 7x} = \bigg \{ \dfrac{\infty}{\infty} \bigg\} = \begin{vmatrix} x \to \infty \\ x + 5x^{2} - x^{3} \sim -x^{3} \\ 2x^{3} - x^{2} + 7x \sim 2x^{3}\end{vmatrix} = \\\\\lim_{x \to \infty} \dfrac{-x^{3}}{2x^{3}} = -\dfrac{1}{2}$

$8) \ \lim_{x \to \infty} \dfrac{x^{3} + x}{x^{4} - 3x^{2} + 1} = \bigg \{ \dfrac{\infty}{\infty} \bigg\} = \begin{vmatrix} x \to \infty \\ x^{3} + x \sim -x^{3} \\ x^{4} - 3x^{2} + 1 \sim x^{4}\end{vmatrix} = \lim_{x \to \infty} \dfrac{-x^{3}}{x^{4}} = \\\\= \lim_{x \to \infty} -\dfrac{1}{x} = -\dfrac{1}{\infty} = 0$

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