We split this into two problems: #\lim_{x\rightarrow\infty}lnx-\lim_{x\rightarrow\infty}\frac{x+2}{x^2+1}# (1). Since the rate of change (derivative) of #\lnx# is equal to #1/x#, which is strictly increasing and is greater than zero on the interval #(0,\infty)#, we can see that #\lnx# is strictly increasing on said interval, and thus that #\lnx# tends to #\infty# as #x# tends to #\infty#.
Now we are left with #\infty + \lim_{x\rightarrow\infty}\frac{x+2}{x^2+1}#. Since it would be helpful not to have so many things that tend to infinity, divide every term by #x^2# to get #\lim_{x\rightarrow\infty}\frac{1/x+2/x^2}{1+1/x^2}#. Since #1/x, 2/x^2, 1/x^2# all tend to zero as #x\rightarrow\infty#, we are left with # \lim_{x\rightarrow\infty}\frac{x+2}{x^2+1}=\frac{0+0}{1+0}=0#. Substituting this into (1), we get #\lim_{x\rightarrow\infty}lnx-\lim_{x\rightarrow\infty}\frac{x+2}{x^2+1}=\infty-0=\infty#.
