Question

Помогите решить предел. С подробным писменым объяснением.
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Answer 1

$6)\; \lim\limits _{x \to 1}(x^4-1)tg\frac{\pi x}{2}=[\; 0\cdot \infty \; ]=\lim\limits _{x \to 1}\frac{x^4-1}{\frac{1}{tg(\pi x/2)}}=\lim\limits _{x \to 1}\frac{x^4-1}{ctg\frac{\pi x}{2}}=[\; Lopital\; ]=\\\\=\lim\limits _{x \to 1}\frac{4x^3}{\frac{-1}{sin^2\frac{\pi x}{2}}\cdot \frac{\pi}{2}}=-\lim\limits _{x \to 1}\frac{8x^3\cdot sin^2\frac{\pi x}{2}}{\pi }=-\frac{8}{\pi }\cdot \lim\limits _{x \to 1}x^3\cdot sin^2\frac{\pi x}{2}=\\\\=-\frac{8}{\pi }\cdot 1^3\cdot 1^2=-\frac{8}{\pi }$

$7)\; \; \lim\limits _{x \to \infty}(1-\frac{1}{x+3})^{7-5x}=\lim\limits _{x \to \infty}\Big ((1-\frac{1}{x+3})^{x+3}\Big )^{\frac{7-5x}{x+3}}=e^{\lim\limits _{x \to \infty}\frac{7-5x}{x+3} }=e^{-5}$

$8)\; \; \lim\limits _{x \to \infty}(\frac{x-5}{x-7})^{\frac{1-3x}{2}}}=\lim\limits _{x \to \infty}\Big ((1+\frac{2}{x-7})^{\frac{x-7}{2}}\Big )^{\frac{2\cdot (1-3x)}{(x-7)\cdot 2}}=e^{\lim\limits _{x \to \infty}\frac{1-3x}{x-7}}=e^{-3}$

$9)\; \; \; \; \; f(x)^{g(x)}=e^{lnf(x)^{g(x)}}=e^{g(x)\cdot lnf(x)}\; \; \Rightarrow \\\\\lim\limits _{x\to x_0}\, f(x)^{g(x)}=\lim\limits _{x \to x_0}\, e^{lnf(x)^{g(x)}}=\lim\limits _{x\to x_0}\, e^{g(x)\cdot lnf(x)}\\\\\\\lim\limits _{x\to 0}\, (sin2x)^{3tg7x}=[\; 0^0\; ]=\lim\limits _{x \to 0}\, e^{3tg7x\cdot lnsin2x}=e^{\lim\limits _{x\to 0}3tg7x\cdot lnsin2x}=\\\\=e^{0}=1$

$\star \; {\lim\limits _{x\to 0}tg7x\cdot lnsin2x=\lim\limits _{x\to 0}\frac{ln(sin2x)}{\frac{1}{tg7x}}=[\, Lopital\, ]=$

$=\lim\limits _{x \to 0}\frac{\frac{2\, cos2x}{sin2x}}{\frac{-1}{tg^27x}\cdot \frac{1}{cos^27x}\cdot 7}=-\frac{2}{7}\cdot \lim\limits _{x \to 0}\frac{ctg2x}{\frac{1}{sin^27x}}=-\frac{2}{7}\cdot \lim\limits _{x \to 0}\frac{\frac{-2}{sin^22x}}{\frac{-14\cdot cos7x}{sin^37x}}=\\\\=-\frac{2}{7}\cdot \frac{2}{14}\cdot \lim\limits _{x \to 0}\frac{sin^37x}{sin^22x\cdot cos7x}=[\, sin7x\sim 7x\; ,\; sin2x\sim 2x\; ,\; cos7x\to 1]=\\\\=-\frac{2}{49}\cdot \lim\limits _{x\to 0}\frac{(7x)^3}{(2x)^2\cdot 1}=-\frac{2}{49}\cdot \lim\limits _{x\to 0}\frac{343x}{4}=-\frac{2}{49}\cdot 0=0$