The discussion revolves around calculating the limit of a complex expression involving a combination of trigonometric, logarithmic, and inverse trigonometric functions as \( x \) approaches 0.
Some participants are actively working through the application of l'Hôpital's rule and sharing their findings. There is a recognition of the limits approaching positive and negative infinity, indicating a complex behavior of the limit as \( x \) approaches 0.
There is an emphasis on the behavior of the logarithmic function and the arcsine function as \( x \) approaches 0, with participants questioning the implications of these behaviors on the limit calculation.
blob84 said:Homework Statement
\lim_{x \rightarrow 0} \frac{(1+tan(3x))^{1/3} x^2}{log_2(arcsin(x^3)+1)}
Homework Equations
The Attempt at a Solution
i tried to use l'hopital twice but i get a very weird function, so there another way to calculate it?
blob84 said:thank you Dick,
i get:
Dx^2= 2x,
D log_{2}(arcsin(x^3)+1)=\frac{3x^2}{log2(arcsin(x^3)+1)\sqrt{1-x^6}},
\lim_{x \to 0} \frac{2x}{\frac{3x^2}{log{2} (arcsin(x^3)+1) \sqrt{1-x^6} } } = <br /> \frac{2}{3}\lim_{x \to 0} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} →<br /> \frac{2}{3}\lim_{x \to 0^-} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x}= \frac{log2}{0^-} = -\infty \vee<br /> \frac{2}{3}\lim_{x \to 0^+} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} = \frac{log2}{0^+}=+\infty<br />