limits problem

limits problem

Postby Guest » Sat Dec 26, 2020 9:48 am

[tex]\lim_{x \to \infty}[x(arctg\frac{x+1}{x+2}-arctg\frac{x}{x+2})][/tex]
Guest
 

Re: limits problem

Postby HallsofIvy » Tue Jan 05, 2021 9:32 am

The limit as x goes to infinity is always difficult since we cannot just set x equal to "infinity".

So divide both numerator and denominator of [tex]\frac{x+ 1}{x+ 2}[/tex] by x to get [tex]\frac{1+ \frac{1}{x}}{1+
\frac{2}{x}}[/tex]. As x goes to infinity, 1/x goes to 0 so that fraction goes to 1. What is arctan(1)? Do the same with [tex]\frac{x}{x+ 2}[/tex] to get [tex]\frac{1}{1+ \frac{2}{x}}[/tex]. That also goes to 1 as x goes to infinity so the two arctangents cancel. That limit is of the form "infinity times 0" which is indeterminate. You might try L'Hopital's rule.

HallsofIvy
 
Posts: 341
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 119


Return to Calculus - integrals, lim, functions



Who is online

Users browsing this forum: No registered users and 1 guest