Q102 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

$\lim _{x \rightarrow 0} \frac{2 \tan x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 \tan x+1}-\sqrt{3 \tan ^2 x+\sin x+1}}=$

  • A
    1
  • B
    3
  • Correct answer
    C
    6
  • D
    $\frac{2}{3}$

Solution

$\lim _{x \rightarrow 0} \frac{2 \tan x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 \tan x+1}-\sqrt{3 \tan ^2 x+\sin x+1}}$ $=\lim _{x \rightarrow 0} \frac{\left(\sqrt{4 \sin ^2 x+2 \tan x+1}+\sqrt{3 \tan ^2 x+\sin x+1}\right)}{\left(4 \sin ^2 x+2 \tan x+1\right)-\left(3 \tan ^2 x+\sin x+1\right)}$ $=\lim _{x \rightarrow 0} \frac{2\left(2 \tan x+x-2 \sin ^2 \frac{x}{2}\right)}{\left(4 \sin ^2 x-3 \tan ^2 x\right)+(2 \tan x-\sin x)}$ $=\lim _{x \rightarrow 0} \frac{2\left(2 \tan x+x-2 \sin ^2 \frac{x}{2}\right)}{\sin ^2 x\left(4-\frac{3}{\cos ^2 x}\right)+\sin x\left(\frac{2}{\cos x}-1\right)}$ $=\lim _{x \rightarrow 0} \frac{2\left(2 \tan x+x-2 \sin ^2 \frac{x}{2}\right)}{\tan ^2 x\left(4 \cos ^2 x-3\right)+\tan x(2-\cos x)}$ $=\lim _{x \rightarrow 0} \frac{2\left(2 \frac{\tan x}{x}+\frac{x}{x}-\sin \frac{x}{2} \cdot \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)}{\tan x \cdot \frac{\tan x}{x}\left(4 \cos ^2 x-3\right)+\frac{\tan x}{x}(2-\cos x)}$ $=\frac{2(2+1)}{1 \cdot(2-1)}=6$.

Hence, the correct answer is option: 'C'.

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