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December 13th, 2016, 10:57 AM   #1
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Question L'Hôpital rule not working correctly? Well, probably it is me who is working wrongly

I am trying to define the limit of the function f(x)=sqrt(x^4+1) / x, when x goes to zero. I tried using the L'Hôpital rule, but I got 0.

$\displaystyle \lim _{x\to 0}\left(\frac{\sqrt{x^4+1}}{x}\right)$

I have done this:

$\displaystyle \lim _{x\to 0}\left(\frac{\frac{1}{2}\left(x^4+1\right)^{-\frac{1}{2}}\cdot 4x^3}{1}\right)$

$\displaystyle x=0$

and got... $\displaystyle \lim _{x\to 0}\left(\frac{\frac{1}{2}\left(x^4+1\right)^{-\frac{1}{2}}\cdot 4x^3}{1}\right)\:\:=0$

Well, I know it is wrong but why the L'Hôpital does not work??? Where have I made a mistake?

Last edited by skipjack; December 13th, 2016 at 03:25 PM.
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December 13th, 2016, 11:12 AM   #2
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L'Hôpital is properly used ONLY when the limits of numerator and denominator individually are both zero or both unbounded. That is not the case in your problem.

$\displaystyle \lim _{x \rightarrow 0} \sqrt{x^4 + 1} = 1 \ne 0.$
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Last edited by skipjack; December 13th, 2016 at 03:25 PM.
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December 13th, 2016, 11:13 AM   #3
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You cannot use L'Hôpital ... direct substitution does not yield the indeterminate form 0/0

fyi, the limit does not exist.
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Last edited by skipjack; December 13th, 2016 at 03:26 PM.
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December 13th, 2016, 11:15 AM   #4
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Thanks, it was quite stupid from me to not think about that.

Last edited by skipjack; December 13th, 2016 at 03:26 PM.
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December 13th, 2016, 02:16 PM   #5
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Originally Posted by srecko View Post
Thanks, it was quite stupid from me to not think about that.
No, just not experienced enough not to slip up every now-and-again. It happens to all of us.

-Dan
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Last edited by skipjack; December 13th, 2016 at 03:26 PM.
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December 13th, 2016, 09:00 PM   #6
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L'Hôpital rule not working correctly?
Blame the user not the tool!
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