2

How to evaluate $$\lim_{x \to 0} \frac{\ln(1 - \sin x) + x} {x^2}$$ without using l'Hôpital? I am not able to substitute the right infinitesimal. Is there a substitute?

Background

  1. We have yet not done Taylor expansions.
  2. I know that $\ln$ around 1 tends to 0 and $\sin$ around 0 tends to 0.
Community
  • 1
Drunk Cat
  • 21
  • Every limit can be computed without using L'Hôpital's rule. – Did Oct 18 '14 at 16:20
  • @Did No different solutions? We have yet not done Taylor. Not for now. – Drunk Cat Oct 18 '14 at 16:23
  • 1
    What methods have you learned? What do you know about $\ln$ around $1$ and $\sin$ around $0$? – quid Oct 18 '14 at 16:25
  • Notable limits, and something about infinitesimal and infinite principles(substitution and elimination). I know that ln around 1 tends to 0 and sin around 0 tendes to 0. – Drunk Cat Oct 18 '14 at 16:29
  • That is good. But I an sorry I do not yet fully understand what you know, and I think you need to know somthing more than this to do that excercice. Do you know something that resembles form $\sin x$ is $x - x^3/3!+ O(x^5)$? Or $\ln (1-x)$ is $-x -x^2/2+ O(x^3)$ where possibly $x$ is an infinitessimal? – quid Oct 18 '14 at 16:52
  • Out of curiosity O(x) is order? I think still have much to learn XD Can you link me some material about this? Thanks for everything ^_^ – Drunk Cat Oct 18 '14 at 17:04
  • You are welcome. Yes, here $O(x^3)$ means that the other terms are at least of order $3$ in $x$. See http://en.wikipedia.org/wiki/Big_O_notation especially the section Infinitesimal asymptotics there. – quid Oct 18 '14 at 17:11

4 Answers4

1

If Series Expansion is allowed,

$$\ln(1-\sin x)=-\sin x-\frac{\sin^2 x}2-O(\sin^3x)$$

Again,$$\sin x=x-O(x^3)$$

lab bhattacharjee
  • 274,582
1

You might use Taylor series expansions: $$\begin{align} \sin x &= x+O(x^3)\\ \ln(1+x)&= x-\frac12x^2+O(x^3)\\ \implies \ln(1-\sin x)+x&=\frac12x^2+O(x^3)\\ \implies \frac{\ln(1-\sin x)+x}{x^2}&=\frac12+O(x)\end{align}$$

Hagen von Eitzen
  • 374,180
1

You need to know the two limits (in addition to the standard limits like $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$): $$\lim_{x \to 0}\frac{x - \log(1 + x)}{x^{2}} = \frac{1}{2},\,\,\lim_{x \to 0}\frac{x - \sin x}{x^{2}} = 0$$ The first of these limits is bit difficult to handle without L'Hospital's Rule and has been calculated in this answer. The second limit is solved in this answer.

Note that if we put $x = -t$ in the first limit we get $$\lim_{t \to 0}\frac{\log(1 - t) + t}{t^{2}} = -\frac{1}{2}$$ and we will use this form here. We proceed as follows $$\begin{aligned}L &= \lim_{x \to 0}\frac{\log(1 - \sin x) + x}{x^{2}}\\ &= \lim_{x \to 0}\frac{\log(1 - \sin x) + \sin x + x - \sin x}{x^{2}}\\ &= \lim_{x \to 0}\frac{\log(1 - \sin x) + \sin x}{x^{2}} + \lim_{x \to 0}\frac{x - \sin x}{x^{2}}\\ &= \lim_{x \to 0}\frac{\log(1 - \sin x) + \sin x}{\sin^{2}x}\cdot\frac{\sin^{2}x}{x^{2}}\\ &= \lim_{x \to 0}\frac{\log(1 - \sin x) + \sin x}{\sin^{2}x}\\ &= \lim_{t \to 0}\frac{\log(1 - t) + t}{t^{2}}\text{ (putting }t = \sin x)\\ &= -\frac{1}{2}\end{aligned}$$

Community
  • 1
Paramanand Singh
  • 87,309
0

Recall that $\ln (1 - x) = -\sum_{i=1}^{\infty} x^n/n$, near $0$. So, $$\ln (1 - \sin x) = -\sum_{i=1}^{\infty} (\sin x)^n/n = - x -x^2/2 + O(x^3).$$

I guess you can finish from there.

quid
  • 42,135