Can I solve limit of natural log without using L'Hospital Rule

Question

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.

Answer 1

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 $x \to 3$.

Answer 2

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 $\lim_{y \to 0} \frac{\ln(1+y)}y=1$.

Answer 3

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, $$ \lim_{x\to0}\frac{\log(1+x)}x=1\tag{5} $$

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