Question
$\mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x} =$
$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{1}{{1 + {x^2}}}}}{1} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{{1 + {x^2}}} = \frac{1}{{1 + 0}} = 1$.
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