Solving indeterminate form of Limit without Using L'hopital's rule

Question

Is there a way to solve $\lim\limits_{x \to 1} \dfrac{\ln(x)}{x-1}$ without using L'hopital?

Answer 1

Following your idea in your comment you may use the well known limit $\lim_{y\to 0}(1+y)^{\frac 1y} = e$ as follows:

Seting $x-1 = y$ you get

$$\lim_{x\to 1}\frac{\ln x}{x-1}= \lim_{y\to 0}\frac{\ln (1+y)}{y}= \lim_{y\to 0}\ln(1+y)^{\frac 1y}= \ln e = 1$$

Answer 2

Rewrite $\ln{(x)}$ as $\ln{\left(1-(1-x)\right)}$. From the taylor series of $\ln{\left(1-x\right)}$ about $x=1$, the numerator can be expressed as $(x-1)+O(x-1)$ for $x \approx 1$. $$\lim_{x \to 1} \frac{x-1+O(x-1)}{x-1}=1$$

Answer 3

If you have learned derivatives, then here's a take on your question:

Since $\ln 1=0$, the limit cam be expressed as $\displaystyle\lim_{x\to 1}\dfrac{\ln x-\ln 1}{x-1}=\left.\dfrac{d}{dx}\ln x\right|_{x=1}=1$, by the definition of a derivative.