Why does L'Hopital not apply to $\lim_{x \to \infty} \frac{5x+3}{\sqrt{4x^2-x+7}}$? I am given the following limit:

$$\lim_{x \to \infty} \frac{5x+3}{\sqrt{4x^2-x+7}}$$

By simply dividing all factors by $x$, one can wind up with the result $\frac{5}{2}$. However, this is also a limit of the form $\frac{\infty}{\infty}$, thus one should be able to apply L'Hopital:

$$\lim_{x \to \infty} \frac{5}{\frac{8x-1}{2\sqrt{4x^2-x+7}}} = 0$$

seeing as we are dividing a constant number by something infinitely large. Where is the mistake in my logic here?

EDIT: Found the mistake after writing this question, will update post with correct answer if this is not already sufficiently answered when I'm finished calculating.
The application of L'Hospital's on radical  seems to be never ending.

$$\text{As }\lim_{x \to \infty} \frac{5x+3}{\sqrt{4x^2-x+7}}=\sqrt{\lim_{x \to \infty} \frac{(5x+3)^2}{4x^2-x+7}},$$

Now applying L'Hospital's
$$\lim_{x \to \infty} \frac{(5x+3)^2}{4x^2-x+7}=\lim_{x \to \infty} \frac{2(5x+3)\cdot5}{8x-1}$$ which is again of the form $\displaystyle\frac\infty\infty$

So applying L'Hospital's again, $$ \lim_{x \to \infty} \frac{10(5x+3)}{8x-1}=\lim_{x \to \infty} \frac{10\cdot5}8=\frac{25}4$$
my answer:

$\lim_{n\to -\infty} \frac{5x}{\sqrt{4x^2-x+7}}=-\frac{5}{2}$

$\lim_{n\to +\infty} \frac{5x}{\sqrt{4x^2-x+7}}=\frac{5}{2}$