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Asphyxiated
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Homework Statement
[tex] \lim_{x \to \infty} \frac {4x}{\sqrt{2x^{2}+1}} [/tex]
Homework Equations
The Attempt at a Solution
So at first glance I saw that this equation ends up at infinity/infinity so I tried to use L'hopital's rule and got:
[tex] \lim_{x \to \infty} \frac {4}{2x(2x^{2}+1)^{-1/2}} [/tex]
That seemed to have sent me in a wrong direction so I went back and just factored out the x^2 from the radical and it worked out such as:
[tex] \lim_{x \to \infty} \frac {4x}{\sqrt{x^{2}(2+1x^{-2})}} [/tex]
[tex] \lim_{x \to \infty} \frac {4x}{x\sqrt{2+1x^{-2}}} [/tex]
[tex] \lim_{x \to \infty} \frac {4}{\sqrt{2+1x^{-2}}} [/tex]
and with the limit applied:
[tex] \lim_{x \to \infty} \frac {4}{\sqrt{2+1x^{-2}}} = \frac {4}{\sqrt{2}} = \sqrt{8}[/tex]
which is correct, so my question is why didn't L'Hopital's Rule work when it was infinity/infinity or would it have worked out and I just stopped too soon?