$\lim_{x \to\ 0} e^4^x-e^-^2^x/2arctgx-sinx$

Question

Математика

Владилен

6 год назад

$\lim_{x \to\ 0} e^4^x-e^-^2^x/2arctgx-sinx$

ОТВЕТЫ

Алексей

Jul 16, 2019

$\lim_{x \to 0} e^{4x} - e^{ \frac{-2x}{2arctg(x) - sin(x)} }amp;#10;amp;#10; \lim_{x \to 0}e^{4x} - \lim_{x \to 0}e^{ \frac{-2x}{2arctg(x) - sin(x)} }amp;#10;amp;#10;1 - \lim_{x \to 0}e^{ \frac{-2x}{2arctg(x) - sin(x)} }amp;#10;amp;#10;1 - e^{\lim_{x \to 0} \frac{-2x}{2arctg(x) - sin(x)} }amp;#10;amp;#10;1 - e^{ \frac{\lim_{x \to 0}-2x}{\lim_{x \to 0}(2arctg(x) - sin(x))} }amp;#10;amp;#10;1 - e^{ -2\frac{-x}{\lim_{x \to 0}(2arctg(x) - \lim_{x \to 0sin(x))} }amp;#10;amp;#10;1 - e^{ -2\frac{-x}{\lim_{x \to 0}(2arctg(x) - x} }amp;#10;amp;#10;1 - e^{ -2\frac{-x}{0 - x}amp;#10;amp;#10;1 - e^{ -2}amp;#10;amp;#10;$

$\lim_{x \to 0} e^{4x} - e^{ \frac{-2x}{2arctg(x) - sin(x)} } = 1 -e^{-2}$

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