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?


  • 2
    $\begingroup$ Please write those images in MathJax. A tutorial can be found here $\endgroup$ Mar 18 '17 at 20:57
  • $\begingroup$ How did you apply L'hopitals, it doesn't seam to help but still doesn't give 0 as result $\endgroup$
    – kingW3
    Mar 18 '17 at 21:00
  • $\begingroup$ I used it multiple times until I ended up with $\frac{16}{5\sqrt{16x^2 + 3}}$ This approaches 0 as x approaches negative infinity $\endgroup$ Mar 18 '17 at 21:02
  • $\begingroup$ @ShoaibAhmed There should be an $x$ in the numerator $\endgroup$ Mar 18 '17 at 21:05
  • $\begingroup$ Ok. My mistake. Sorry for wasting your time. $\endgroup$ Mar 18 '17 at 21:14

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.

  • $\begingroup$ Thank You for the reply. But how do I know that this approaches -0.8? If I substitute a large negative number for x, I can see that the numerator approaches negative infinity, and the denominator approaches positive infinity. This would prompt me to use L'hopital again. Am I missing something? $\endgroup$ Mar 18 '17 at 21:07
  • $\begingroup$ @ShoaibAhmed You, like many students, seem to think every limit needs to be attacked with LHopital, but that simply won't work here. You get suck in a loop, as the other answers have noted. You are required to use a different technique $\endgroup$ Mar 18 '17 at 21:08
  • $\begingroup$ OK. Thanks. I was just wondering why I got stuck in a loop. I thought maybe I was doing something wrong. $\endgroup$ Mar 18 '17 at 21:10
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    $\begingroup$ @ShoaibAhmed You got stuck in a loop for a very similar reason to the reason why you'd get stuck in a loop trying to evaluate $\lim_{x \to \infty} \frac{\exp(x)}{\exp(x)}$ using L'Hôpital's rule. $\endgroup$ Mar 18 '17 at 21:11

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.

  • $\begingroup$ Thank You. That is a useful strategy $\endgroup$ Mar 18 '17 at 21:11
  • $\begingroup$ @ShoaibAhmed It helps minimizing the risks when you consider $\sqrt{a^2}$; since we use $t>0$, you can write $\sqrt{t^2}=t$. $\endgroup$
    – egreg
    Mar 18 '17 at 21:20

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.

  • $\begingroup$ Thank You. I got to that point above after using L'hopital twice. However, I made the mistake of using it a third time, and somehow ended up with the limit of 0. I might've made a mistake there. Thanks $\endgroup$ Mar 18 '17 at 21:13

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