The value of lim-x-x - tan-1 - x-1-x-2 -tan-1 - x-x-2 - -is1-1-1-2-1-2-

The correct option is C 1/2 =lim_x→∞ x [ tan^ 1 ( x+1/x+2 ) tan^ 1 ( x/x+2 ) ]is =lim_x→∞ x tan^ 1 ( x+1/x+2 x/x+2/1+x+1/x+2x/x+2) =lim_x→∞ x tan^ 1 ( x+2/2x ...

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Standard XII

Mathematics

Standard Limits to Remove Indeterminate Form

The value of ...

Question

The value of ${lim}_{x \to \infty} x [t a n^{- 1} (\frac{x + 1}{x + 2}) - t a n^{- 1} (\frac{x}{x + 2})] i s$

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Solution

The correct option is C
1/2

$= {lim}_{x \to \infty} x [t a n^{- 1} (\frac{x + 1}{x + 2}) - t a n^{- 1} (\frac{x}{x + 2})] i s = {lim}_{x \to \infty} x t a n^{- 1} (\frac{\frac{x + 1}{x + 2} - \frac{x}{x + 2}}{1 + \frac{x + 1}{x + 2} \frac{x}{x + 2}})$

$= {lim}_{x \to \infty} x t a n^{- 1} (\frac{x + 2}{2 x^{2} + 5 x + 4}) = {lim}_{x \to \infty} ⎡ ⎢ ⎣ \frac{t a n^{- 1 (\frac{x + 2}{2 x^{2} + 5 x + 4})}}{\frac{x + 2}{2 x^{2} + 5 x + 4}} ⎤ ⎥ ⎦ \times \frac{x (x + 2)}{2 x^{2} + 5 x + 4}$
Using standard limits -
$= 1 \times \frac{1}{2} = \frac{1}{2}$

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Similar questions

Q. Solve the following equations for x:
(i)

(ii)
(iii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x
(iv) tan⁻¹

(\frac{1 - x}{1 + x}) - \frac{1}{2}

tan⁻¹x = 0, where x > 0
(v) cot⁻¹x − cot⁻¹(x + 2) =

\frac{π}{12}

, x > 0
(vi) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹

(\frac{8}{79})

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(vii)

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(ix)

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(i) tan⁻¹2x + tan⁻¹3x = nπ +

\frac{3 π}{4}

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹

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tan⁻¹x = 0, where x > 0
(v) cot⁻¹x − cot⁻¹(x + 2) =

\frac{π}{12}

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\tan^{- 1} (\frac{x - 2}{x - 4}) + \tan^{- 1} (\frac{x + 2}{x + 4}) = \frac{π}{4}

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\tan^{- 1} (2 + x) + \tan^{- 1} (2 - x) = \tan^{- 1} \frac{2}{3}, where x < - \sqrt{3} or, x > \sqrt{3}

(x)

\tan^{- 1} \frac{x - 2}{x - 1} + \tan^{- 1} \frac{x + 2}{x + 1} = \frac{π}{4}

Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ + $\frac{3 π}{4}$

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹ $\frac{8}{31}$

(iii) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(iv) sin⁻¹x + sin⁻¹2x = $\frac{π}{3}$

(v) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(vi) $\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin $(\cot^{- 1} \frac{1}{2})$

(ix) tan⁻¹ $(\frac{1 - x}{1 + x}) - \frac{1}{2}$ tan⁻¹x = 0, where x > 0

(x) cot⁻¹x − cot⁻¹(x + 2) = $\frac{π}{12}$ , x > 0

(xi) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹ $(\frac{8}{79})$ , x > 0

(xiii) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$

Q. The value of

lim x \to \infty x [{tan}^{- 1} \frac{x + 1}{x + 2} - {cot}^{- 1} \frac{x + 2}{x}]

is:

Q. Assertion :

{cosec}^{- 1} (3 / 2) + {cos}^{- 1} (2 / 3) - 2 {cot}^{- 1} (1 / 7) - {cot}^{- 1} 7

is equal to

{cot}^{- 1} 7

Reason:

{sin}^{- 1} x + {cos}^{- 1} x = π / 2, {tan}^{- 1} x + {cot}^{- 1} x = π / 2, {cosec}^{- 1} x = {sin}^{- 1} (1 / x),

{cot}^{- 1} (1 / x) = {tan}^{- 1} (x), {cot}^{- 1} (x) = {tan}^{- 1} (1 / x)

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The value of limx→∞ x[tan−1(x+1x+2)−tan−1(xx+2)]is

The correct option is C 1/2 =limx→∞ x[tan−1(x+1x+2)−tan−1(xx+2)]is=limx→∞ x tan−1(x+1x+2−xx+21+x+1x+2xx+2) =limx→∞ x tan−1(x+22x2+5x+4)=limx→∞⎡⎢⎣tan−1(x+22x2+5x+4)x+22x2+5x+4⎤⎥⎦×x(x+2)2x2+5x+4 Using standard limits - =1×12=12

The value of ${lim}_{x \to \infty} x [t a n^{- 1} (\frac{x + 1}{x + 2}) - t a n^{- 1} (\frac{x}{x + 2})] i s$