The value of limx- 01xtan-1x-12x-1-4 is equal to-

The correct option is C 12 lim_x→ #10;0(tan^ 1(x+12x+1) tan^ 11x)(00) =lim_x→ #10; #10;0tan^ 1((x+12x+1) 11+(x+12x+1))x =( )lim_x→ ...

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Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

The value of ...

Question

The value of $lim x \to 0 \frac{1}{x} ({tan}^{- 1} (\frac{x + 1}{2 x + 1}) - \frac{π}{4})$ is equal to.

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\frac{- 1}{2}

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Solution

The correct option is C $\frac{- 1}{2}$
$lim x \to 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{{tan}^{- 1} (\frac{x + 1}{2 x + 1}) - {tan}^{- 1} 1}{x} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ (\frac{0}{0})$
$= lim x \to 0 \frac{{tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{(\frac{x + 1}{2 x + 1}) - 1}{1 + (\frac{x + 1}{2 x + 1})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠}{x}$
$= (-) lim x \to 0 \frac{{tan}^{- 1} (\frac{- x}{3 x + 2})}{(\frac{- x}{3 x + 2}) \cdot (3 x + 2)} = \frac{- 1}{2}$ .

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The value of limx→01x(tan−1(x+12x+1)−π4) is equal to.

The correct option is C −12limx→0⎛⎜ ⎜ ⎜ ⎜⎝tan−1(x+12x+1)−tan−11x⎞⎟ ⎟ ⎟ ⎟⎠(00)=limx→0tan−1⎛⎜ ⎜ ⎜ ⎜⎝(x+12x+1)−11+(x+12x+1)⎞⎟ ⎟ ⎟ ⎟⎠x=(−)limx→0tan−1(−x3x+2)(−x3x+2)⋅(3x+2)=−12.

The value of $lim x \to 0 \frac{1}{x} ({tan}^{- 1} (\frac{x + 1}{2 x + 1}) - \frac{π}{4})$ is equal to.