# limits problem

### limits problem

$$\lim_{x \to \infty}[x(arctg\frac{x+1}{x+2}-arctg\frac{x}{x+2})]$$
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### Re: limits problem

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 $$\frac{x+ 1}{x+ 2}$$ by x to get $$\frac{1+ \frac{1}{x}}{1+ \frac{2}{x}}$$. As x goes to infinity, 1/x goes to 0 so that fraction goes to 1. What is arctan(1)? Do the same with $$\frac{x}{x+ 2}$$ to get $$\frac{1}{1+ \frac{2}{x}}$$. 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

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