Calculate limits without l'Hôpital I am trying to calculate the following limits
$$ \lim_{x\to\infty}(2x+1) \ln \left(\frac{x-3}{x+2}\right) $$
and
$$ \lim_{x\to1}\frac{x}{3x-3}\ln(7-6x) $$
I can't use l'Hôpital's rule, so I am not sure how to solve it.
By continuity of the logarithm,
$$ \lim_{x\to\infty} \ln \left(\frac{x-3}{x+2}\right)^{2x+1}
= \lim_{x\to\infty} \ln \left(1-\frac5{x+2}\right)^{2(x+2)-3}\\
=\ln((e^{-5})^2)-3\lim_{x\to\infty}\ln \left(1-\frac5{x+2}\right).$$
$$\lim_{x\rightarrow \infty }\left ( 2x+1 \right )\ln \left ( \frac{x-3}{x+2} \right )=\lim_{x\rightarrow \infty }\frac{\frac{\mathrm{d} }{\mathrm{d} x}\ln \left ( \frac{x-3}{x+2} \right )}{\frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{\left ( 2x+1 \right )}}=-\lim_{x\rightarrow \infty }\frac{5\left ( 2x+1 \right )^2}{2\left ( x-3 \right )\left ( x+2 \right )}=-\frac{5}{2}\lim_{x\rightarrow \infty }\frac{\left ( 2x+1 \right )^2}{\left ( x-3 \right )\left ( x+2 \right )}=-\frac{5}{2}\cdot 4\lim_{x\rightarrow \infty }\frac{2x+1}{2x-1}=-\frac{5}{2}\cdot 4=-10$$
$$\lim_{x\rightarrow 1 }\frac{x}{3x-3}\ln \left ( 7-6x \right )=\lim_{x\rightarrow 1 }\frac{\ln \left ( 7-6x \right )}{3x-3}=\lim_{x\rightarrow 1 }\frac{\frac{\mathrm{d} }{\mathrm{d} x}\ln \left ( 7-6x \right )}{\frac{\mathrm{d} }{\mathrm{d} x}3x-3}=\lim_{x\rightarrow 1 }\frac{2}{6x-7}=-2$$