Question

$\underset{x\to 1}{\lim }{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1-\cos \left(x-1\right)}{{\left(x-1\right)}^2}}$

Answer 1

$$\begin{gathered} \underset{x\to 1}{\lim }{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1-\cos \left(x-1\right)}{{\left(x-1\right)}^2}} \\ =\underset{x\to 1}{\lim }{\left\{1+\frac{x^3+2x^2+x+1}{x^2+2x+3}-1\right\}}^{\frac{1-\cos \left(x-1\right)}{{\left(x-1\right)}^2}} \\ =e^{\underset{x\to 1}{\lim }\left\{\frac{x^3+2x^2+x+1-x^2-2x-3}{x^2+2x+3}\right\}\times \left[\frac{1-\cos \left(x-1\right)}{{\left(x-1\right)}^2}\right]} \\ =e^{\underset{x\to 1}{\lim }\left\{\frac{x^3+x^2-x-2}{x^2+2x+3}\right\}\times \left[\frac{2{\sin }^2{\displaystyle \frac{\left(x-1\right)}{2}}}{4\times {\displaystyle \frac{{\left(x-1\right)}^2}{4}}}\right]} \\ =e^{\left(\frac{1+1-1-2}{1+2+3}\right)\times \frac{1}{2}} \\ =e^{-\frac{1}{12}} \end{gathered}$$

Disclaimer: The solution given in the book is incorrect. However, the solution given here has been created according to the question given in the book.