How can I evaluatie this limit at infinity?

Question

I have to evaluate $$\lim_{x \rightarrow \infty} \left( \sqrt{4x-1} - \sqrt{9x} \right),$$ and I cannot use L'Hopital's Rule. At first I thought it should be zero, but I decided to do it analytically.

The first thing I do is multiply and divide by $\sqrt{4x-1} + \sqrt{9x}$ and I get $$\lim_{x \rightarrow \infty} \frac{-5x - 1}{\sqrt{4x-1} + \sqrt{9x}}.$$

The problem here is that $-5x-1 \rightarrow - \infty$ and $\sqrt{4x-1} + \sqrt{9x} \rightarrow \infty,$ so I would have something like $\frac{- \infty}{\infty}.$ It is not clear to me how to proceed, so I would like to receive some help to complete the problem.

Answer 1

Next step: Divide numerator and denominator by $x$. Then look at the limit again.

Answer 2

Note that from here you can conclude as follow

$$\frac{-5x - 1}{\sqrt{4x-1} + \sqrt{9x}}=\frac{x}{\sqrt x}\,\frac{-5 - 1/x}{\sqrt{4-1/x} + \sqrt{9}}=\sqrt x\,\frac{-5 - 1/x}{\sqrt{4-1/x} + \sqrt{9}}\to+\infty\cdot\frac{-5}5=-\infty$$

Answer 3

$\sqrt{4x-1}$ is a bit less than $\sqrt{4x}=2\sqrt x$ while $\sqrt{9x}$ actually is $3\sqrt x$. Thus $\sqrt{4x-1}-\sqrt{9x}<2\sqrt x-3\sqrt x=-\sqrt x $. As $-\sqrt x\to-\infty$ as $x\to\infty$ then $\sqrt{4x-1}-\sqrt{9x}\to-\infty$ too.