lim x cos( 1 + ecos(x) + 73x4) x→0 x
时间: 2023-05-29 17:01:43 浏览: 53
This is an indeterminate form of the type 0/0.
Using L'Hopital's rule:
lim x cos(1 - ecos(x) + 73x^4) / x
= lim x (-sin(1 - ecos(x) + 73x^4)) * (-esin(x) - 292x^3) / 1 (applying L'Hopital's rule)
= (-sin(1)) * (-esin(0) - 0^3) (as the limit of the numerator and denominator goes to zero)
= sin(1)
Therefore, the limit is sin(1).
相关问题
x+ln(x+1)等价无穷小是多少
当$x\rightarrow 0$时,有:
$x\ln(x+1) = \frac{\ln(x+1)}{\frac{1}{x}}$
由洛必达法则可得:
$$
\lim_{x\rightarrow 0} \frac{\ln(x+1)}{\frac{1}{x}} = \lim_{x\rightarrow 0} \frac{\frac{1}{x+1}}{-\frac{1}{x^2}} = \lim_{x\rightarrow 0} -\frac{x^2}{x+1} = 0
$$
因此,$x\ln(x+1)$是$x$趋于$0$时的等价无穷小是$0$。
X趋于0+时sin^2x/x^2等于多少
当 $x$ 趋近于 $0$ 时,$\sin^2 x/x^2$ 的极限可以用洛必达法则求解:
$$\lim_{x\rightarrow 0^+} \frac{\sin^2 x}{x^2} = \lim_{x\rightarrow 0^+} \frac{2\sin x \cos x}{2x} = \lim_{x\rightarrow 0^+} \frac{\cos x}{1} = 1$$
因此,当 $x$ 趋近于 $0$ 时,$\sin^2 x/x^2$ 的极限为 $1$。