Can I not use L'hopital rule here?

Question

The question is:

$$ \lim_{x\to-\infty}\frac{\sqrt{16x^2+3}}{5x-2} $$

The given solution is:

This makes sense since as $x$ approaches negative infinity, we look at the left side of the $\sqrt{x^2}$ function, which is $-x$.

I tried using L'hopital rule before looking at the answer and got that the limit was $0$. Am I not allowed to use L'hopital here? I think I am. As $x$ approaches negative infinity, the numerator approaches infinity, and the denominator approaches negative infinity. We want to see which approaches their limit faster, so we look at their rates of change, i.e. their derivatives. So I fugured we could use L'hopital here.

Am I misunderstanding something?

Thanks

Answer 1

You seem to have done your calculations incorrectly. You should get $\displaystyle\lim_{x \to -\infty} \frac{16x}{5\sqrt{16x^2+3}}$ and not $\displaystyle\lim_{x \to -\infty} \frac{16}{5\sqrt{16x^2+3}}$ as you state in the comments

---

We start with $$\lim_{x \to -\infty} \frac{\sqrt{16x^2+3}}{5x-2}$$ Taking the derivative of the numerator, we get $\frac{16x}{5\sqrt{16x^2+3}}$
Taking the derivative of the denominator, we simply get $5$

---

Combining these, we transform our limit into $$\lim_{x \to -\infty} \frac{16x}{5\sqrt{16x^2+3}}$$ Which in fact ALSO approaches $-0.8$ as $x$ approaches $-\infty$, though I am not convinced this is a simpler limit to solve.

Answer 2

You can certainly try l’Hôpital, with the problem it leads nowhere: the quotient of the derivatives is $$ \frac{16x}{5\sqrt{16x^2+3}} $$ and a further application will lead to $$ \frac{\sqrt{16x^2+3}}{5x} $$ Now the loop goes on ad infinitum.

---

A simpler strategy is to set $t=-1/x$, so you get $$ \lim_{t\to0^+}\frac{\sqrt{\dfrac{16}{t^2}+3}}{-\dfrac{5}{t}-2}= \lim_{t\to0^+}\frac{\sqrt{16+3t^2}}{-5-2t} $$ that poses no challenge.

Answer 3

It does work.

$$\lim_{x \to -\infty} \frac{\sqrt{16x^2+3}}{5x+2} = \lim_{x \to -\infty} \frac{16x (16x^2+3)^{-1/2}}{5} = \lim_{x \to -\infty} \frac{16x}{5 \sqrt{16x^2+3}}$$ L'Hôpital again: $$\lim_{x \to -\infty} \frac{16 \sqrt{16x^2+3}}{5 \times 16x}$$ which reduces to $$\frac{1}{5} \lim_{x \to -\infty} \frac{\sqrt{16x^2+3}}{x}$$

This is easier: it's got rid of the annoying $+2$ from the denominator, and perhaps it makes it clearer that the next step is to bring the $\frac{1}{x}$ inside the square root. As others note, you can't proceed using L'Hôpital's rule here without just repeating your earlier work, but this form is more amenable to the "do the obvious" technique.