# Math Help - l'Hôpital's rule

1. ## l'Hôpital's rule

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

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

3. 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

4. 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.