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I want to evaluate the following limit:

$$\lim_{x\to 1^{+}} \frac{\ln(x)}{x-1}$$

After some thought, I was able to recognize this as the derivative of $\ln(x)$ evaluated at $x=1$ yielding $1$ as the value of the limit.

I would like to know other methods of evaluating this limit.

Thanks for your help.

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Did you try write $ln$ as a series? –  DiegoMath Aug 12 '13 at 16:18
    
Perhaps I should have mentioned that I'm a beginning calculus student. However, I'll look into it. Thanks. –  Gorg Aug 12 '13 at 16:22
    
One can compare the slopes of two functions(Which is indirectly L'Hospital's rule). –  Inceptio Aug 12 '13 at 16:26
2  
Your method is good. I guess you have not yet reached L'Hospital's Rule. Nor have you reached series, but if you make the substitution $x=1+t$ you are looking at the limit as $t$ approaches $0$ of $\frac{\ln(1+t)}{t}$ and then you can expand $\ln(1+t)$ in a power series. –  André Nicolas Aug 12 '13 at 16:26
    
You can use L'hopital's rule. –  Mhenni Benghorbal Aug 12 '13 at 20:54

3 Answers 3

up vote 0 down vote accepted

Remember that $$ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^n=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$ for $|x|\leq1$, unless $x=-1$.

Hence,

$$\begin{matrix}\lim_{x\to1^+}\frac{ln(x)}{x-1}&=&\lim_{x\to1+}\frac{(x-1)-\frac{(x-1)^2}{2}+\cdots}{x-1}\\ &=&\lim_{x\to1+}\left(1-\frac{x-1}{2}+\frac{(x-1)^2}{3}-\cdots\right)\\ &=&1\end{matrix}$$

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Substitute $x-1=h$ where $h\to 0^+$. Transforming the limit in terms of $h$, you get

$$\lim_{h\to0^+} \frac{\ln(1+h)}{h}$$

You get the limit in the form of a fundamental limit, which equates to $1$.

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I tried to substitute a simple parameter, given that $\mathcal{D}:(\,0\,;1)\cup(1;+\infty)$: $$ \ln (x)=z\to x=e^z $$ then the fundamental limit is shown: $$ \lim_{x\to 1^+}z=0^+\to\lim_{z\to 0^+}\frac{z}{e^z-1}=\left(\lim_{z\to 0^+}\frac{e^z-1}{z}\right)^{-1}=1 $$

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