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.
 A: 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}
A: METHODOLOGY $1$: Non-Calculus Based
We can evaluate the limit 
$$\lim_{x\to \infty}\left(x\log(16x+14)-x\log(16x+7)\right)=\lim_{x\to \infty}\left(x\log\left(1+\frac{7}{8x}\right)-x\log\left(1+\frac{7}{16x}\right)\right)\tag 1$$
using only a standard inequality and the Squeeze Theorem.
In THIS ANSWER, and THIS ONE, I showed that the logarithm function satisfies the inequalities
$$\frac{x}{1+x}\le \log (1+x)\le x \tag 2$$
using only the limit definition of the exponential function and Bernoulli's Inequality.  
Applying $(2)$ to the right-hand side of $(1)$ reveals
$$x\left(\frac{\frac7{8x}}{1+\frac7{8x}}\right)-x\frac{7}{16x}\le x\log\left(1+\frac{7}{8x}\right)-x\log\left(1+\frac{7}{16x}\right)\le x\left(\frac{7}{8x}\right)-x\left(\frac{\frac7{16x}}{1+\frac7{16x}}\right)$$
which simplifies to 
$$\frac7{16}\left(\frac{1-\frac{7}{8x}}{1+\frac{7}{16x}}\right)\le x\log\left(1+\frac{7}{8x}\right)-x\log\left(1+\frac{7}{16x}\right)\le \frac7{16}\left(\frac{1+\frac{7}{8x}}{1+\frac{7}{16x}}\right) \tag 3$$
Finally, applying the Squeeze Theorem to $(3)$ yields the result
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\left(x\log(16x+14)-x\log(16x+7)\right)=\frac{7}{16}}$$

METHODOLOGY $2$:  Series Expansion 
If one wishes to use a series expansion, we can write
$$\begin{align}
\left(x\log\left(1+\frac{7}{8x}\right)-x\log\left(1+\frac{7}{16x}\right)\right)&=\left(\frac78 +O\left(\frac1x\right)\right)-\left(\frac7{16}+O\left(\frac1x\right)\right)\\\\
&=\frac7{16}+O\left(\frac1x\right)\\\\
&\to \frac{7}{16}\,\,\text{as}\,\,x\to \infty
\end{align}$$
and we are done!
