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

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

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Standard XI

Mathematics

Equations with Solutions at Boundary Values

The value of ...

Question

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

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Solution

The correct option is C
1/2

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

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

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

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

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