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Ответ:

Предел:

[tex]\boldsymbol{\boxed{ \lim_{x \to \infty} \bigg ( \frac{5 - x}{1 - x} \bigg )^{3x - 1} =e^{-12}}}[/tex]

Примечание:

Теорема:

[tex]\boxed{\displaystyle \lim_{x \to a } u(x)^{v(x)} = [1]^{\infty} \Longrightarrow \lim_{x \to a } u(x)^{v(x)} = e^\bigg {\displaystyle \lim_{x \to a } (v(x)(u(x) - 1)) }}[/tex]

Пошаговое объяснение:

[tex]\displaystyle \lim_{x \to \infty} \bigg ( \frac{5 - x}{1 - x} \bigg )^{3x - 1} = \lim_{x \to \infty} \bigg ( \frac{4 + 1 - x}{1 - x} \bigg )^{3x - 1} = \lim_{x \to \infty} \bigg ( \frac{4}{1 - x} + \frac{1 - x}{1 - x} \bigg )^{3x - 1} =[/tex]

[tex]\displaystyle = \lim_{x \to \infty} \bigg ( \frac{4}{1 - x} + 1 \bigg )^{3x - 1} = [1]^{\infty} = e^\bigg{\displaystyle \lim_{x \to \infty} \Bigg ( (3x - 1)\bigg (\frac{4}{1 - x} + 1 -1 \bigg ) \Bigg )} =[/tex]

[tex]= e^\bigg{\displaystyle \lim_{x \to \infty} \frac{4(3x - 1)}{1 - x} } = e^\bigg{\displaystyle 4\lim_{x \to \infty} \frac{3x - 1}{1 - x} } = e^\bigg{\displaystyle -4\lim_{x \to \infty} \frac{1 - 3x }{1 - x} } =[/tex]

[tex]= e^\bigg{\displaystyle -4\lim_{x \to \infty} \frac{1 - x - 2x }{1 - x} } = e^\bigg{\displaystyle -4\lim_{x \to \infty} \bigg ( \frac{1 - x}{1 - x} - \frac{2x }{1 - x} \bigg )} } = e^\bigg{\displaystyle -4\lim_{x \to \infty} \bigg (1 +\frac{-2x }{1 - x} \bigg )} } =[/tex]

[tex]= e^\bigg{\displaystyle (-4) \cdot 2\lim_{x \to \infty} \bigg (\frac{1}{2} +\frac{-x }{1 - x} \bigg )} } = e^\bigg{\displaystyle -8\lim_{x \to \infty} \bigg (\frac{1}{2} +\frac{1-x - 1 }{1 - x} \bigg )} } =[/tex]

[tex]= e^\bigg{\displaystyle -8\lim_{x \to \infty} \bigg (\frac{1}{2} +\frac{1-x }{1 - x} -\frac{1 }{1 - x} \bigg )} } = e^\bigg{\displaystyle -8\lim_{x \to \infty} \bigg (\frac{1}{2} +1 -\frac{1 }{1 - x} \bigg )} } =[/tex]

[tex]= e^\bigg{\displaystyle -8\lim_{x \to \infty} \bigg (1,5 -\frac{1 }{1 - x} \bigg )} } = e^\bigg {-8 \cdot 1,5} = e^\bigg{-12}[/tex].