Toán Giúp tớ câu này với: Tìm lim->0 (ln(cos(x))/x^2 17/07/2021 By Maria Giúp tớ câu này với: Tìm lim->0 (ln(cos(x))/x^2
Đáp án: \(\lim\limits_{x\to 0}\dfrac{\ln(\cos x)}{x^2}=-\dfrac12\) Giải thích các bước giải: \(\begin{array}{l}\quad \lim\limits_{x\to 0}\dfrac{\ln(\cos x)}{x^2}\\= \lim\limits_{x\to 0}\dfrac{-\dfrac{\sin x}{\cos x}}{2x}\qquad (l’Hôpital)\\= -\dfrac12\lim\limits_{x\to 0}\dfrac{\sin x}{x}\cdot\lim\limits_{x\to 0}\dfrac{1}{\cos x}\\= -\dfrac12\cdot 1\cdot \dfrac{1}{\cos 0}\\= -\dfrac12\end{array}\) Trả lời
`~rai~` \(\lim\limits_{x\to0}\dfrac{In(cosx)}{x^2}\\=\lim\limits_{x\to 0}\dfrac{In(cosx)’}{(x^2)’}\text{(quy tắc l’hospital)}\\=\lim\limits_{x\to 0}\dfrac{\dfrac{(cosx)’}{cosx}}{2x}\\=\lim\limits_{x\to 0}\dfrac{\dfrac{-sinx}{cosx}}{2x}\\=-\dfrac{1}{2}.\lim\limits_{x\to 0}\dfrac{sinx}{x}.\lim\limits_{x\to0}\dfrac{1}{cosx}\\=-\dfrac{1}{2}.1.\dfrac{1}{cos0}\\=-\dfrac{1}{2}.\) $# Băng Cướp Cầu Vồng.$ Trả lời
Đáp án:
\(\lim\limits_{x\to 0}\dfrac{\ln(\cos x)}{x^2}=-\dfrac12\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to 0}\dfrac{\ln(\cos x)}{x^2}\\
= \lim\limits_{x\to 0}\dfrac{-\dfrac{\sin x}{\cos x}}{2x}\qquad (l’Hôpital)\\
= -\dfrac12\lim\limits_{x\to 0}\dfrac{\sin x}{x}\cdot\lim\limits_{x\to 0}\dfrac{1}{\cos x}\\
= -\dfrac12\cdot 1\cdot \dfrac{1}{\cos 0}\\= -\dfrac12
\end{array}\)
`~rai~`
\(\lim\limits_{x\to0}\dfrac{In(cosx)}{x^2}\\=\lim\limits_{x\to 0}\dfrac{In(cosx)’}{(x^2)’}\text{(quy tắc l’hospital)}\\=\lim\limits_{x\to 0}\dfrac{\dfrac{(cosx)’}{cosx}}{2x}\\=\lim\limits_{x\to 0}\dfrac{\dfrac{-sinx}{cosx}}{2x}\\=-\dfrac{1}{2}.\lim\limits_{x\to 0}\dfrac{sinx}{x}.\lim\limits_{x\to0}\dfrac{1}{cosx}\\=-\dfrac{1}{2}.1.\dfrac{1}{cos0}\\=-\dfrac{1}{2}.\)
$# Băng Cướp Cầu Vồng.$