The original question ask to evaluate the limit below:

$$ \lim_{x \to \infty}{x\sqrt{1-\frac{1}{x}}} $$

From what I conclude this equation can't have L'Hospital rule applied and this limit just goes to infinity. Am I correct or am I missing something?

  • $\begingroup$ What are the hypotheses under which one can apply De L'Hopital's rule? $\endgroup$ – Riccardo Orlando Nov 4 '16 at 18:08
  • $\begingroup$ @riccardoorlando Can you clarify what you mean? $\endgroup$ – Cameron Wetzel Nov 4 '16 at 18:09
  • $\begingroup$ Rewrite it as $\frac{x\sqrt{x-1}}{\sqrt{x}}$. Then you can apply L'Hopital, and get infinity as the result. $\endgroup$ – Thomas Andrews Nov 4 '16 at 18:10
  • $\begingroup$ It's not indeterminate. $\endgroup$ – T.J. Gaffney Nov 4 '16 at 18:12
  • $\begingroup$ @CameronWetzel De L'Hopital's rule can only apply when the function is of the form $\frac00$ or $\frac \infty \infty$ $\endgroup$ – Riccardo Orlando Nov 4 '16 at 18:12

L'Hopitals rule

Classicly you would apply L'Hopitals rule when you have $\lim_\limits{x\to a} \frac {f(x)}{g(x)}$ with $\lim_\limits{x\to a} f(x) = \lim_\limits{x\to a} g(x) = 0.$ In wich case $\lim_\limits{x\to a} \frac {f(x)}{g(x)} =\lim_\limits{x\to a} \frac {f'(x)}{g'(x)}$

Now there are some cases where you might have

$\lim_\limits{x\to a} f(x)g(x)$ with $\lim_\limits{x\to a} f(x) = 0$ and $\lim_\limits{x\to a} g(a) = \infty$ and you could also apply L'Hopitals rule as you are quick tranformation away from $\lim_\limits{x\to a} f(x)g(x) = \lim_\limits{x\to a} \frac {f(x)}{\frac{1}{g(x)}}.$

But if you had: $\lim {x\to\infty} \sqrt{x+a} - \sqrt{x+b}$ you could not apply L'Hopitals with the functions in that form. You need to make it into a product or a ratio.

Pre-amble complete:

$f(x) = x, g(x) = \sqrt{1-\frac 1x}$

$\lim_\limits{x\to\infty} f(x) = \infty\\ \lim_\limits{x\to\infty} g(x) = 1$

Even if you make it into a ratio as above you are going to get $\frac {1}{0}$ and not $\frac 00$. Can't (or no need to) apply L'Hoptials, the limit is going to equal $\infty$

As has been pointed out in the comments, it is possible to change the set up.

$f(x) = x\sqrt{x-1}, g(x) = \frac 1{\sqrt x}$ and then you would get the necessary indeterminate. But that is overkill.

  • $\begingroup$ Thank you. I eventually did evaluate it and kept get answers that only gave $\infty$ so I just determined it evaluated to that $\endgroup$ – Cameron Wetzel Nov 4 '16 at 18:38

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