\displaystyle{\left(\text{lim}\right)}_{{{x}\vec{\infty}}}\text{{}{x}+{5}\text{)}{{\tan}^{{-{1}}}{\left({x}+{5}\right)}}-{\left({x}+{1}\right)}{{\tan}^{{-{1}}}{\left({x}+{1}\right)}}{\rbrace} is equal to $π$ (b) $2 π$ (c) $\frac{π}{2}$ (d) none of these

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\displaystyle{\left(\text{lim}\right)}_{{{x}\vec{\infty}}}\text{{}{x}+{5}\text{)}{{\tan}^{{-{1}}}{\left({x}+{5}\right)}}-{\left({x}+{1}\right)}{{\tan}^{{-{1}}}{\left({x}+{1}\right)}}{\rbrace} is equal to

π

(b)

2 π

(c)

\frac{π}{2}

(d) none of these

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Question Text

π

(b)

2 π

(c)

\frac{π}{2}

(d) none of these

\displaystyle{\left(\text{lim}\right)}_{{{x}\vec{\infty}}}\text{{}{x}+{5}\text{)}{{\tan}^{{-{1}}}{\left({x}+{5}\right)}}-{\left({x}+{1}\right)}{{\tan}^{{-{1}}}{\left({x}+{1}\right)}}{\rbrace} is equal to $π$ (b) $2 π$ (c) $\frac{π}{2}$ (d) none of these

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\displaystyle{\left(\text{lim}\right)}_{{{x}\vec{\infty}}}\text{{}{x}+{5}\text{)}{{\tan}^{{-{1}}}{\left({x}+{5}\right)}}-{\left({x}+{1}\right)}{{\tan}^{{-{1}}}{\left({x}+{1}\right)}}{\rbrace} is equal to π (b) 2π (c) 2π​ (d) none of these

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Practice questions from similar books

\displaystyle{\left(\text{lim}\right)}_{{{x}\vec{\infty}}}\text{{}{x}+{5}\text{)}{{\tan}^{{-{1}}}{\left({x}+{5}\right)}}-{\left({x}+{1}\right)}{{\tan}^{{-{1}}}{\left({x}+{1}\right)}}{\rbrace} is equal to $π$ (b) $2 π$ (c) $\frac{π}{2}$ (d) none of these