Question no 11 and 12

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Dear student,

Q n o 11 p u t x = 1 + t s o t h a t a s x \to 1 t \to 0 \underset{x \to 1}{l = l i m} \frac{1 + \ln x - x}{1 - 2 x + x^{2}} = \underset{t \to 0}{l i m} \frac{1 + \ln (1 + t) - 1 - t}{1 - 2 (1 + t) + (1 + t)^{2}} = \underset{t \to 0}{l i m} \frac{\ln (1 + t) - t}{t^{2}} U \sin g \ln (1 + t) = t - \frac{t^{2}}{2} + \frac{t^{3}}{3} + . . . w e g e t l = \underset{t \to 0}{l i m} \frac{t - \frac{t^{2}}{2} + \frac{t^{3}}{3} + . . . - t}{t^{2}} = - \frac{t^{2} / 2 + H i g h e r o r d e r t e r m s}{t^{2}} = \frac{- 1}{2}

Q n o 12 W e k n o w - 1 \leq \cos x \leq 1 s o t h a t \frac{- 1}{2 x + 1} \leq \frac{1 - 2 \cos x}{2 x + 1} \leq \frac{3}{2 x + 1} w e k n o w \underset{x \to \infty}{l i m} \frac{- 1}{2 x + 1} = 0 a n d \underset{x \to \infty}{l i m} \frac{3}{2 x + 1} = 0 H e n c e b y s a n d w i c h t h e o r e m \underset{x \to \infty}{l i m} \frac{1 - 2 \cos x}{2 x + 1} = 0 H o p e t h a t h e l p s R e g a r d s