Calculating the Limit of a Complex Expression

[blob84]

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?

 

[Dick]

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?

I'd be interested to see these weird functions. l'Hopital is the best way to do it. There is a trick that might help you. lim x->0 (1+tan(3x))^(1/3)=1. That's pretty harmless. I wouldn't include it in the l'Hopital treatment. In general if a factor of your function has a nonzero finite limit, just set it aside and concentrate on the rest.

 

[blob84]

thank you Dick,
i get:
$$Dx^2= 2x$$,
$$D log_{2}(arcsin(x^3)+1)=\frac{3x^2}{log2(arcsin(x^3)+1)\sqrt{1-x^6}}$$,
$$\lim_{x \to 0} \frac{2x}{\frac{3x^2}{log{2} (arcsin(x^3)+1) \sqrt{1-x^6} } } = <br /> \frac{2}{3}\lim_{x \to 0} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} →<br /> \frac{2}{3}\lim_{x \to 0^-} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x}= \frac{log2}{0^-} = -\infty \vee<br /> \frac{2}{3}\lim_{x \to 0^+} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} = \frac{log2}{0^+}=+\infty$$

 

[Dick]

thank you Dick,
i get:
$$Dx^2= 2x$$,
$$D log_{2}(arcsin(x^3)+1)=\frac{3x^2}{log2(arcsin(x^3)+1)\sqrt{1-x^6}}$$,
$$\lim_{x \to 0} \frac{2x}{\frac{3x^2}{log{2} (arcsin(x^3)+1) \sqrt{1-x^6} } } = <br /> \frac{2}{3}\lim_{x \to 0} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} →<br /> \frac{2}{3}\lim_{x \to 0^-} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x}= \frac{log2}{0^-} = -\infty \vee<br /> \frac{2}{3}\lim_{x \to 0^+} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} = \frac{log2}{0^+}=+\infty$$

Looks ok to me!