Evaluate - x- 0 Lt - 1 - sin 2 x - - 1 - 2 tan x sin x - tan 2 x

lim_x→ 0(1+sin^2x)^1/3 (1 2tan x)^1/4sin x+tan^2 x Applying L Hospital law =lim_x→013(1+sin^2x)^ 2/3(sin 2x) 14(1 2tan x)^ 3/4( 2sec^2x)cos x+2tan xsec ...

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Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

Evaluate : ...

Question

Evaluate : $L t x \to 0 \frac{\sqrt[3]{1 + {sin}^{2} x} - \sqrt[4]{(1 - 2 tan x)}}{sin x + {tan}^{2} x}$

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Solution

$l i m_{x \to 0} \frac{(1 + {sin}^{2} x)^{1 / 3} - (1 - 2 t a n x)^{1 / 4}}{sin x + t a n^{2} x}$
Applying L-Hospital law
$= l i m_{x \to 0} \frac{\frac{1}{3} (1 + {sin}^{2} x)^{- 2 / 3} (sin 2 x) - \frac{1}{4} (1 - 2 t a n x)^{- 3 / 4} (- 2 s e c^{2} x)}{cos x + 2 t a n x s e c^{2} x} = \frac{0 - \frac{1}{4} (- 2)}{1 + 0} = \frac{1}{2}$

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Evaluate : Ltx→03√1+sin2x−4√(1−2tanx)sinx+tan2x

limx→0(1+sin2x)1/3−(1−2tanx)1/4sinx+tan2xApplying L-Hospital law=limx→013(1+sin2x)−2/3(sin2x)−14(1−2tanx)−3/4(−2sec2x)cosx+2tanxsec2x=0−14(−2)1+0=12

Evaluate : $L t x \to 0 \frac{\sqrt[3]{1 + {sin}^{2} x} - \sqrt[4]{(1 - 2 tan x)}}{sin x + {tan}^{2} x}$