Why can't l'hospital's rule be used in this example?

Aug 2018
3
0
Czech Republic
1576424199844.png
I tried using the rule here, but can never get the correct answer (3). Why can't I use the rule in this example even though i get infinity/infinity after substitution?
 

skeeter

Math Team
Jul 2011
3,210
1,733
Texas
Because using L'Hôpital cycles as a never ending $\dfrac{\infty}{\infty}$ form ...

here are three iterations, note what happens ...

$\displaystyle \lim_{x \to \infty} \dfrac{3x-2}{\sqrt{x^2-4}-2}$

$\displaystyle \lim_{x \to \infty} \dfrac{3\sqrt{x^2-4}}{x}$

$\displaystyle \lim_{x \to \infty} \dfrac{3x}{\sqrt{x^2-4}}$

$\displaystyle \lim_{x \to \infty} \dfrac{3\sqrt{x^2-4}}{x}$

... and so on.

dividing numerator and denominator by $x$ ...

$\displaystyle \lim_{x \to \infty} \dfrac{\frac{3x}{x}-\frac{2}{x}}{\frac{\sqrt{x^2-4}}{\sqrt{x^2}}-\frac{2}{x}}$

$\displaystyle \lim_{x \to \infty} \dfrac{3-\frac{2}{x}}{\sqrt{1-\frac{4}{x^2}}-\frac{2}{x}} = 3$
 

mathman

Forum Staff
May 2007
6,892
759
Simpler method: Let $y=\frac{1}{x}$. Then the question becomes $\lim_{y\to 0} \frac{3-2y}{\sqrt{1-4y^2}-2y}=3$. L'Hopital's rule not needed.
 

greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
\(\displaystyle \lim_{x \to \infty} \dfrac{3x-2}{\sqrt{x^2-4}-2}\sim3x/x=3\)
 
  • Like
Reactions: idontknow