the limit mathop limlimitsx to infty x tan 1 x 1x 2 tan 1

The limit $$\mathop {\lim}\limits_{x \to \infty } x\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right) - {{\tan }^{ - 1}}\left( {\frac{x}{{x + 2}}} \right)} \right]$$ is equal to

A. $$2$$

B. $$\frac{1}{2}$$

C. $$ - \frac{1}{3}$$

D. None of these

Answer : $$\frac{1}{2}$$

Solution :
$$\eqalign{ & \mathop {\lim}\limits_{x \to \infty } x\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right) - {{\tan }^{ - 1}}\left( {\frac{x}{{x + 2}}} \right)} \right] \cr & = \mathop {\lim }\limits_{x \to \infty } \,x\,{\tan ^{ - 1}}\left( {\frac{{\frac{{x + 1}}{{x + 2}} - \frac{x}{{x + 2}}}}{{1 + \frac{{x + 1}}{{x + 2}} \cdot \frac{x}{{x + 2}}}}} \right) \cr & = \mathop {\lim }\limits_{x \to \infty } x\,{\tan ^{ - 1}}\left( {\frac{{x + 2}}{{2{x^2} + 5x + 4}}} \right) \cr & = \mathop {\lim }\limits_{x \to \infty } x\left( {\frac{{{{\tan }^{ - 1}}\left( {\frac{{x + 2}}{{2{x^2} + 5x + 4}}} \right)}}{{\frac{{x + 2}}{{2{x^2} + 5x + 4}}}}} \right) \times \frac{{x\left( {x + 2} \right)}}{{2{x^2} + 5x + 4}} \cr & = 1 \times \frac{1}{2} = \frac{1}{2} \cr} $$

Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

Releted Question 1

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$ is

A. $$\frac{6}{{17}}$$

B. $$\frac{7}{{16}}$$

C. $$\frac{16}{{7}}$$

D. none

View Answer

Releted Question 2

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$ is

A. $$\frac{{\sqrt {29} }}{3}$$

B. $$\frac{{29}}{3}$$

C. $$\frac{{\sqrt {3}}}{29}$$

D. $$\frac{{3}}{29}$$

View Answer

Releted Question 3

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$ is

A. zero

B. one

C. two

D. infinite

View Answer

Releted Question 4

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

A. $$ \frac{1}{2}$$

B. 1

C. $$ - \frac{1}{2}$$

D. $$- 1$$

View Answer

The limit $$\mathop {\lim}\limits_{x \to \infty } x\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right) - {{\tan }^{ - 1}}\left( {\frac{x}{{x + 2}}} \right)} \right]$$ is equal to

Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$ is

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$ is

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$ is

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

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Inverse Trigonometry Function

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The limit $$\mathop {\lim}\limits_{x \to \infty } x\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right) - {{\tan }^{ - 1}}\left( {\frac{x}{{x + 2}}} \right)} \right]$$ is equal to

Releted MCQ Question on Trigonometry >> Inverse Trigonometry Function

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$ is

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$ is

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$ is

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

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Releted MCQ Question on
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