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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.

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**

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