Can I solve limit of natural log without using L'Hospital Rule I want to find the following limit: $$\lim_{x\to 2} \dfrac{\ln(x+3)-\ln(5)}{(x-2)}$$
but can we find the limit without using L'Hospitals Rule?
It is an indeterminate form but I do not know how to rewrite it.
 A: We have
$$\lim_{x \to 2} \dfrac{\ln(x+3)-\ln(5)}{x-2} = \lim_{h \to 0} \dfrac{\ln(5+h)-\ln(5)}{h} = \lim_{h \to0} \dfrac{\ln(1+h/5)}h = \dfrac15$$
where we made use of the fact that $\lim_{y \to 0} \frac{\ln(1+y)}y=1$.
A: One way is to realize that by defintion this limit is the derivative of the function $f(x)=\ln(x+3)$ at $x=2$. Assuming that you made a typo for $x \to 3$.
A: Bernoulli's Inequality says that for $n\ge1$ and $x\ge-n$,
$$
1+x\le\left(1+\frac{x}{n}\right)^n\tag{1}
$$
Thus, for all $x$, we have
$$
1+x\le\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x\tag{2}
$$
Substituting $x\mapsto-\frac{x}{1+x}$ gives $\frac1{1+x}\le e^{-\frac{x}{1+x}}$, which implies $e^{\frac{x}{1+x}}\le1+x$.
Thus, for $x\gt-1$,
$$
\frac{x}{1+x}\le\log(1+x)\le x\tag{3}
$$
Dividing by $x$ yields, for $x\gtrless0$,
$$
\frac1{1+x}\lessgtr\frac{\log(1+x)}x\lessgtr1\tag{4}
$$
Thus, by the Squeeze Theorem,
$$
\lim_{x\to0}\frac{\log(1+x)}x=1\tag{5}
$$

Thus, because $\frac{x-2}5\to0$ as $x\to2$,
$$
\begin{align}
\lim_{x\to2}\frac{\log(x+3)-\log(5)}{x-2}
&=\frac15\lim_{x\to2}\frac{\log\left(1+\frac{x-2}{5}\right)}{\frac{x-2}5}\\
&=\frac15\tag{6}
\end{align}
$$
