$$\lim_{x \to \infty} x\bigg(\ln (4x+7)-\ln (4x+8) \bigg)$$
Here's my approach
Let $4x+7=u$ then we have,
$$\lim_{u \to \infty} \frac{7-u}{4} \ln \bigg(\frac{u+1}{u}\bigg)$$
$$=\lim_{u \to \infty} -\frac{1}{4}u \ln \bigg(1+\frac{1}{u}\bigg)$$
$$=-\frac{1}{4}$$
I'm interested other techniques to refresh myself and strengthen my limit toolbox.
Since $\ln$ and $f(x)=e^x$ are continuous functions, we obtain: $$\lim\limits_{x\rightarrow\infty}x(\ln(4x+7)-\ln(4x+8))=\lim\limits_{x\rightarrow\infty}\ln\left(1-\frac{1}{4x+8}\right)^{-(4x+8)\cdot\left(-\frac{x}{4x+8}\right)}=\ln{e^{-\frac{1}{4}}}=-\frac{1}{4}$$
One a bit different approach (with more limited scope): By the Lagrange theorem ($f(a)-f(b)=f'(c)(a-b)$ for some $c\in(a,b)$) $\ln(4x+7)-\ln(4x+8)=\dfrac1{4x+7+\xi}\cdot(-1)$, where $\xi\in(0,1)$. Hence $$ \lim_{x\to+\infty}x(\ln(4x+7)-\ln(4x+8))=-\lim_{x\to+\infty}\frac x{4x+7+\xi}=-\lim_{x\to+\infty}\frac 1{4+7/x+\xi/x}=-\frac14. $$
Or, good old L'Hospital rule: $$ \lim_{x\to+\infty}x(\ln(4x+7)-\ln(4x+8))=\lim_{x\to+\infty}\frac{\ln(4x+7)-\ln(4x+8)}{1/x}=\lim_{x\to+\infty}\frac{4/(4x+7)-4/(4x+8)}{-1/x^2}=\lim_{x\to+\infty}\frac{4/(4x+7)(4x+8)}{-1/x^2}=-\lim_{x\to+\infty}\frac{4x^2}{(4x+7)(4x+8)}=-\frac14. $$
Taylors:
$$\lim x\ln \frac{4x+7}{4x+8}= \lim x \ln (1-\frac 1{4x+8}) =_{x\to \infty} \lim -x\frac 1{4x+8} = -\frac 14$$
Set $1/x= h$
$\log(4x+a)=\ln4x+\ln(1+a/4x)$
$\ln(1+a/4x)=\ln(1+ah/4)$
Finally, $\lim_{h\to0}\dfrac{\ln(1+h)}h=1$