A=3, B=1, C=4, вычислить

Question 1

Question 2

$1)\quad \lim\limits _{x \to \infty} \frac{3x^3+x+1}{4x^3-6x^2+9} = \lim\limits _{x \to \infty}\frac{x^3\cdot \left (3+\frac{1}{x^2}+\frac{1}{x^3}\right )}{x^3\cdot \left (4-\frac{6}{x}+\frac{9}{x^3}\right )} =\frac{3}{4}\\\\2)\quad \lim\limits _{x \to 0} \frac{e^{3arcsin^3x}-1}{ln(1+tg^3(4x))} =[\; (e^{ \alpha }-1)\approx \alpha \; ;\; \; ln( 1+\alpha )\approx \alpha \; ,\; \alpha \to 0]=\\\\= \lim\limits _{x \to 0} \frac{3arcsin^3x}{tg^3(4x)} =[\, arcsin \alpha \approx \alpha \; ,\; tg \alpha \approx \alpha \; ]=$

$= \lim\limits _{x \to 0} \frac{3\cdot x^3}{(4x)^3} = \lim\limits _{x \to 0} \frac{3x^3}{64x^3} =\frac{3}{64}\\\\3)\quad \lim\limits _{x \to +\infty} (7^{x}+x^4)^{ \frac{1}{x}}=7\\\\y(x)=(7^{x}+x^4)^{\frac{1}{x}}\; \; \to \; \; \; lny(x)=\frac{1}{x}\cdot ln(7^{x}+x^4)\; ;\\\\\lim\limits_{x\to +\infty }lny= \lim\limits _{x \to +\infty} \frac{ln(7^{x}+x^4)}{x} =[\frac{\infty }{\infty }\; ]=[\; Lopital\; ]=$

$= \lim\limits _{x \to +\infty} \frac{7^{x}ln7+4x^3}{7^{x}+x^4}=\lim\limits _{x\to +\infty }\frac{7^{x}ln^27+12x^2}{7^{x}ln7+4x^3}$ =

$= \lim\limits _{x \to +\infty} \frac{7^{x}ln^37+24x}{7^{x}ln^27+12x^2}= \lim\limits _{x \to +\infty} \frac{7^{x}ln^47+24}{7^{x}ln^37+24x} = \lim\limits _{x \to +\infty} \frac{7^{x}ln^57}{7^{x}ln^47+24} =\\\\= \lim\limits _{x \to +\infty} \frac{7^{x}ln^67}{7^{x}ln^57} =ln7\\\\lim\, (lny(x)\, )=ln\, (limy(x)\, )=ln7\; \; \to \; \; \lim\limits _{x\to +\infty }y(x)=7$