Find $\lim_{x\to\infty} (x \ln(16x +14) - (x \ln(16x +7))$ using Maclaurin series. I am trying to find the limit of   $\lim_{x\to\infty} (x \ln(16x +14)  - (x \ln(16x +7))$.
I know I have to use Maclaurin series, but something went wrong.

You don't need any series expansion or L'Hospital to find this limit. You need only the continuity of $\ln(x)$ and some arithmetics:
\begin{gather}\lim_{x \to \infty} x\ln (16x+14) - x\ln(16x+7) =\\= \lim_{x \to \infty} x\ln\left({{16x+14}\over{16x+7}}\right) = \lim_{x \to \infty} \ln\left(\left({{16x+14}\over{16x+7}}\right)^x\right) =\\ 
= \lim_{x \to \infty} \ln\left(\left({{16x+14}\over{16x+7}}\right)^x\right) = \ln\left(\lim_{x \to \infty}\left({{16x+14}\over{16x+7}}\right)^x\right) =\\= \ln\left(\lim_{x \to \infty}\left({{1+{14\over {16x}}}\over{1+{7\over {16x}}}}\right)^x\right) =  \ln\left({{e^{14 \over 16}}\over{e^{7\over 16}}}\right)
= \ln\big({e^{7\over 16}}\big)  =  {7\over 16} \end{gather}
