Môn Toán Lớp 12 Giúp mình 2 câu này vớiiii )):

Question

Môn Toán Lớp 12 Giúp mình 2 câu này vớiiii )): Giúp em bài này với ạ em cần gấp, đừng copy nguồn trên mạng nha. Em xin cảm ơn thầy cô và các bạn nhiều.

in progress 0
2 tháng 2022-10-04T16:50:18+00:00 1 Answer 0 views 0

Trả lời ( )

1. $$\begin{array}{l} \text{Câu 4:}\\ a)\quad \lim\limits_{x\to 0}\dfrac{\ln(\cos x)}{x^2}\\ = \lim\limits_{x\to 0}\dfrac{-\dfrac{\sin x}{\cos x}}{2x}\qquad \text{(l’Hôpital)}\\ = \lim\limits_{x\to 0}\left(-\dfrac{1}{2\cos x}\right)\cdot\lim\limits_{x\to 0}\dfrac{\sin x}{x}\\ = -\dfrac{1}{2\cos 0}\cdot 1\\ = – \dfrac12\\ b)\quad \lim\limits_{x\to 0^+}(1 + \sin2021x)^{\tfrac2x}\\ = \lim\limits_{x\to 0^+}e^{\ln(1 + \sin2021x)^{\tfrac2x}}\\ = \lim\limits_{x\to 0^+}e^{\tfrac2x\ln(1 + \sin2021x)}\\ = e^{\lim\limits_{x\to 0^+}\left[\tfrac2x\ln(1 + \sin2021x)\right]}\\ = e^{\lim\limits_{x\to 0^+}\tfrac{2\ln(1 + \sin2021x)}{x}}\\ = e^{\lim\limits_{x\to 0^+}\tfrac{4042\cos2021x}{1 + \sin2021x}}\qquad \text{(l’Hôpital)}\\\\ = e^{\tfrac{4042\cos(2021.0)}{1 + \sin(2021.0)}}\\ = e^{4042}\\ \text{Câu 5:}\\ a)\quad f(x) = x + \sqrt{2x+1}\\ +)\quad f(0) = 1\\ +)\quad f'(x) = 1 + \dfrac{1}{\sqrt{2x + 1}} \Rightarrow f'(0) = 2\\ +)\quad f”(x) = – \dfrac{1}{\sqrt{(2x+1)^3}} \Rightarrow f”(0) = -1\\ +)\quad f”'(x) = \dfrac{3}{\sqrt{(2x+1)^5}} \Rightarrow f”'(0) = 3\\ +)\quad f^{(4)}(x) = -\dfrac{15}{\sqrt{(2x + 1)^7}} \Rightarrow f^{(4)}(0) = – 15\\ \text{Ta được:}\\ f(x)\approx 1 + 2x – \dfrac{x^2}{2}+\dfrac{x^3}{2} – \dfrac{5x^4}{8} + o(x^4)\\ b)\quad f(x) = \ln(x+3)\\ +)\quad f(0) = \ln3\\ +)\quad f'(x) = \dfrac{1}{x+3} \Rightarrow f'(0) = \dfrac{1}{3}\\ +)\quad f”(x) = -\dfrac{1}{(x+3)^2} \Rightarrow f”(0) = -\dfrac{1}{9}\\ +)\quad f”'(x) = \dfrac{2}{(x+3)^3} \Rightarrow f”'(0) = \dfrac{2}{27}\\ +)\quad f^{(4)}(x) = – \dfrac{6}{(x+3)^4} \Rightarrow f^{(4)}(0) = – \dfrac{2}{27}\\ \text{Ta được:}\\ f(x)\approx \ln3 + \dfrac{x}{3} – \dfrac{x^2}{18} + \dfrac{x^3}{81} – \dfrac{x^4}{324} + o(x^4) \end{array}$$