Explanation

(b)
Let $f(x) = \int_{0}^{x}{\left\lbrack \frac{1}{\sqrt{1 + t^{2}}} - \frac{1}{1 + t} \right\rbrack dt}$
$= \int_{0}^{x}{\frac{1}{\sqrt{1 + t^{2}}}dt - \int_{0}^{x}{\frac{1}{1 + t}dt}}$
Let $I_{1} = \int_{0}^{x}{\frac{1}{\sqrt{1 + t^{2}}}dt}$
Put $t = \tan{\theta \Rightarrow dt = \sec^{2}{\theta\ d\theta}}$
$\therefore I_{1} = \int_{0}^{\tan^{- 1}x}\frac{\sec^{- 1}\theta}{\sec\theta}d\theta$
$= \lbrack log(sec\ \theta + \tan{\theta)}\ \rbrack\ _{0}^{\tan^{- 1}x}\ \ \$
$= log(\sec{\tan^{- 1}{x + x)}}$
$\therefore f(x) = log(\sec{\tan^{- 1}{x + x) - \log{(1 + x)}}}$
$\therefore\lim_{x \rightarrow \infty}{f(x)} = \lim_{x \rightarrow \infty}{\log\left( \frac{\sec{tan^{- 1}}x + x}{1 + x} \right)}$
${= \lim_{x \rightarrow \infty}}{\log\left( \frac{\sqrt{1 + x^{2}} + x}{1 + x} \right)}$
$= \log\underset{x \rightarrow \infty}{\ \ lim}{\ \left( \frac{\sqrt{\frac{1}{x^{2}} + 1 + 1}}{\frac{1}{x} + 1} \right)}$
$= \log_{e}2$