Find $\lim\limits_{x\to 0}{\frac{x+\ln (1-x)}{x^2}}$ without derivatives

Question

If we know that $$\lim\limits_{x\to 1}{\frac{\ln x}{x-1}}=1$$ and $$\lim\limits_{x\to 0}{\frac{x+\ln (1-x)}{x^2}}=l\in\mathbb{R}$$ find the value of $l$ without derivatives.

I have only managed to find that: $$\lim\limits_{x\to 0}{\frac{\ln (1-x)}{x}}=-1$$ Any hint?

score 12 · Accepted Answer · answered Sep 20 '15 at 12:33

12

Change $x\to 0$ to $(-x)\to 0$ to get $$ \lim_{x\to 0}\frac{-x+\ln(1+x)}{x^2}=l. $$
Add two limits $$ 2l=l+l=\lim_{x\to 0}\Bigl(\frac{x+\ln(1-x)}{x^2}+\frac{-x+\ln(1+x)}{x^2}\Bigr)= \lim_{x\to 0}\frac{\ln(1-x^2)}{x^2}. $$
Change $x^2=t\to 0$ to get $$ 2l=\lim_{t\to 0}\frac{\ln(1-t)}{t}=-1. $$ Thus, $l=-\frac12$.

answered Sep 20 '15 at 12:33

A.Γ.

29,518

1

Fantastic! Thank you for sharing :). – MrYouMath Sep 20 '15 at 12:36
Nice. How to guarantee that limits exist? – Sep 20 '15 at 12:37
2

@JohnMa The question itself guaranteed... – Asydot Sep 20 '15 at 12:38
@Asydot : This is really a general question. – Sep 20 '15 at 12:41
1

@JohnMa As I understand it is given that the limit exists and is equal to $l$. – A.Γ. Sep 20 '15 at 12:46
@A.G. Yes, I understand that. Your proof is very nice. I was asking how one would know $\ell$ exist. – Sep 20 '15 at 12:50
From the hypothesis, $l\in\mathbb{R}$ so it exists. Great proof!!! – Sep 20 '15 at 12:55
you can apply Hopital rule check this and see "show step by step":http://www.wolframalpha.com/input/?i=lim%28%5Cfrac%7Bx%2Bln%281-x%29%7D%7Bx%C2%B2%7D%2Cx%5Cto+0 – zeraoulia rafik Sep 20 '15 at 14:06

Find $\lim\limits_{x\to 0}{\frac{x+\ln (1-x)}{x^2}}$ without derivatives

1 Answers1

Linked