Hi,how can i solve by l'Hôpital's rule

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- Mar 24th 2011, 01:57 AMBAHADEENl'Hôpital's rule
Hi,how can i solve by l'Hôpital's rule

- Mar 24th 2011, 02:29 AMgirdav
Do you really have to solve it by l'Hôpital rule ? You can write $\displaystyle \sqrt{9x^2+1} =3x\sqrt{1+\frac 1{9x^2}}$ and $\displaystyle \sqrt{4x^2+1} =2x\sqrt{1+\frac 1{4x^2}}$.

- Mar 24th 2011, 02:35 AMBAHADEEN
- Mar 24th 2011, 06:14 AMHallsofIvy
The derivative of $\displaystyle (9x^2+ 1)^{1/2}$ is

$\displaystyle \frac{1}{2}(9x^2+ 1)^{-1/2}(18x)= \frac{9x}{\sqrt{9x^2+ 1}}$

The derivative of $\displaystyle (4x^2+ 1)^{1/2}$ is

$\displaystyle \frac{1}{2}(4x^2+ 1}^{-1/2}(8x)= \frac{4x}{\sqrt{4x^2+ 1}}$

So the ratio of the derivatives is

$\displaystyle \frac{9x}{\sqrt{9x^2+ 1}}\frac{\sqrt{4x^2+ x}}{4x}= \frac{9}{4}\frac{\sqrt{4x^2+ 1}}{\sqrt{9x^2+ 1}}$

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. - Mar 24th 2011, 06:25 AMPlato