# l'Hôpital's rule

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• March 24th 2011, 01:57 AM
BAHADEEN
l'Hôpital's rule
Hi,how can i solve by l'Hôpital's rule
• March 24th 2011, 02:29 AM
girdav
Do you really have to solve it by l'Hôpital rule ? You can write $\sqrt{9x^2+1} =3x\sqrt{1+\frac 1{9x^2}}$ and $\sqrt{4x^2+1} =2x\sqrt{1+\frac 1{4x^2}}$.
• March 24th 2011, 02:35 AM
BAHADEEN
Quote:

Originally Posted by girdav
Do you really have to solve it by l'Hôpital rule ? You can write $\sqrt{9x^2+1} =3x\sqrt{1+\frac 1{9x^2}}$ and $\sqrt{4x^2+1} =2x\sqrt{1+\frac 1{4x^2}}$.

thanks , yes please by l'Hôpital rule
• March 24th 2011, 06:14 AM
HallsofIvy
The derivative of $(9x^2+ 1)^{1/2}$ is
$\frac{1}{2}(9x^2+ 1)^{-1/2}(18x)= \frac{9x}{\sqrt{9x^2+ 1}}$

The derivative of $(4x^2+ 1)^{1/2}$ is
$\frac{1}{2}(4x^2+ 1}^{-1/2}(8x)= \frac{4x}{\sqrt{4x^2+ 1}}$

So the ratio of the derivatives is
$\frac{9x}{\sqrt{9x^2+ 1}}\frac{\sqrt{4x^2+ x}}{4x}= \frac{9}{4}\frac{\sqrt{4x^2+ 1}}{\sqrt{9x^2+ 1}}$

But that is just 9/4 times the reciprocal of the original fraction. If, by L'Hopital's rule, we must have the limit of the original fraction equal to the limit of that, we must have, calling the limit of the original fraction L, L= (9/4)(1/L). Can you solve that for L? Of course, technically, you should also prove that L'Hopital's rule actually works here- that the limit exists.
• March 24th 2011, 06:25 AM
Plato
Quote:

Originally Posted by BAHADEEN
thanks , yes please by l'Hôpital rule

Why would anyone use l'Hôpital's rule on that limit?
What a waste of time, let alone discouraging understanding.