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Calculate limit

  • Thread starter blob84
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  • #1
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Homework Statement



[tex]\lim_{x \rightarrow 0} \frac{(1+tan(3x))^{1/3} x^2}{log_2(arcsin(x^3)+1)}[/tex]

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?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement



[tex]\lim_{x \rightarrow 0} \frac{(1+tan(3x))^{1/3} x^2}{log_2(arcsin(x^3)+1)}[/tex]

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.
 
  • #3
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thank you Dick,
i get:
[tex]Dx^2= 2x[/tex],
[tex]D log_{2}(arcsin(x^3)+1)=\frac{3x^2}{log2(arcsin(x^3)+1)\sqrt{1-x^6}}[/tex],
[tex]\lim_{x \to 0} \frac{2x}{\frac{3x^2}{log{2} (arcsin(x^3)+1) \sqrt{1-x^6} } } =
\frac{2}{3}\lim_{x \to 0} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} →
\frac{2}{3}\lim_{x \to 0^-} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x}= \frac{log2}{0^-} = -\infty \vee
\frac{2}{3}\lim_{x \to 0^+} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} = \frac{log2}{0^+}=+\infty
[/tex]
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
thank you Dick,
i get:
[tex]Dx^2= 2x[/tex],
[tex]D log_{2}(arcsin(x^3)+1)=\frac{3x^2}{log2(arcsin(x^3)+1)\sqrt{1-x^6}}[/tex],
[tex]\lim_{x \to 0} \frac{2x}{\frac{3x^2}{log{2} (arcsin(x^3)+1) \sqrt{1-x^6} } } =
\frac{2}{3}\lim_{x \to 0} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} →
\frac{2}{3}\lim_{x \to 0^-} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x}= \frac{log2}{0^-} = -\infty \vee
\frac{2}{3}\lim_{x \to 0^+} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} = \frac{log2}{0^+}=+\infty
[/tex]
Looks ok to me!
 

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