The value of limx- x - tan-1 x-1-x-2 -tan-1 x-x-2 -is

=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^2+5x+4 ) =lim_x→∞ [ tan^ 1 ...

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Standard XIII

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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1/2

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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. The value of

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

is:

Q. Solve the following equations for x:
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\frac{3 π}{4}

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

\frac{8}{31}

(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) =

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

Q. If

t a n^{- 1} (\frac{x - 1}{x - 2}) + t a n^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{2}

then find the value of x

Q. If

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

. Find the value of x.

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})

, x > 0
(vii)

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

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