Only tell me the answer

1.	Only tell me the answer
Answer» ANSWER:- $lt \: (x \: -> \: <klux>0</klux>) = \frac{cos {}^{2} - <klux>SIN</klux> {}^{2}x \: - 1 }{ \sqrt{x {}^{2} + 1 - 1 } } \\ lt \: (x \: -> \: 0) = \frac{cos2x \: - 1}{ \sqrt{x {}^{2} + 1 } - 1}$ .°. 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, ${\huge{\boxed{\tt{\red{-2}}}}}$

Answer»