Using L'Hopitals rule I have done: $$\lim_{x \to \infty} \frac{1}{\sqrt{1-x^2}}$$ Then diving top and bottom by $x$: $$\lim_{x \to \infty} \frac{\frac1x}{\sqrt{\frac1{x^2}-1}}$$ To me this seems to suggest that the limit is zero. However, when I use graphing software to check, it shows that it tends to infinity. What have I done wrong?
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6$\arcsin x$ does not makes sense when $x\not\in[-1,1]$. Ditto for $\frac{1}{\sqrt{1-x^2}}$. – Martín-Blas Pérez Pinilla Dec 12 '14 at 11:49
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If complex values are allowed, L'Hôpital is forbidden: http://math.stackexchange.com/questions/363339/lh%C3%B4pitals-rule-for-vector-valued-functions. – Martín-Blas Pérez Pinilla Dec 12 '14 at 11:54
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What can i do instead? – RobChem Dec 12 '14 at 12:15
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1The limit does not exist because the function is not even defined for large $x$. – André Nicolas Dec 12 '14 at 12:52