Question

Only tell me the answer

Answer 1

ANSWER:-

$lt \: (x \: -> \: 0) = \frac{cos {}^{2} - sin {}^{2}x \: - 1 }{ \sqrt{x {}^{2} + 1 - 1 } } \\ lt \: (x \: -> \: 0) = \frac{cos2x \: - 1}{ \sqrt{x {}^{2} + 1 } - 1}$

.°. cos$\theta$ -1 / √1 -1 =0/0 Form.

L'HOSPITAL RULE,

$lt \: (x \: -> \: 0) = \: \frac{ - sin2x \: (2) - 0}{ \frac{1}{ \cancel1 \sqrt{x {}^{2} + 1 \: . \cancel1x - 0} } } \\ lt \: (x \: -> \: 0) = \: 2(\frac{sin2x}{1x} ) \times ( \sqrt{x {}^{2} + 1 }) \times 1$

.°. lt ( $\theta$ -> 0) = sin$\theta$ /$\theta$ =1

$lt \: (x \: -> \: 0) = - 2 \times 1 \times \sqrt{x {}^{2} + 1} \\ .°. \: lt \: (x \: -> \: 0) = \: - 2 \times \sqrt{0 + 1} \\ .°. \: lt \: (x \: -> \: 0) = \: - 2 \times 1 \\ .°. \: lt \: (x \: -> \: 0) = \red {- 2}$

.°. So final answer is,

${\huge{\boxed{\tt{\red{-2}}}}}$