Answer
$\infty$
Work Step by Step
The given limit can be written:
$$\lim_{x \to \infty}\ln{\frac{1+x^2}{1+x}}$$
$$=\lim_{x \to \infty}\ln{(x-\frac{x-1}{1+x})}$$
$$=\lim_{x \to \infty}\ln{\left(x-1+\frac{2}{1+x}\right)}$$
$$=\lim_{x \to \infty}\ln{\left(x-1+\frac{2}{1+x}\right)}$$
Since $\displaystyle\lim_{x \to \infty}\left(x-1+\frac{2}{1+x} \right)=\infty$
so:
$$\lim_{x \to \infty}\ln{\left(x-1+\frac{2}{1+x}\right)}=\infty$$
