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I am trying to compute the following limit without L'Hôpital's rule : $$L=\lim_{x\to 0}\frac{\ln(\tan(x)+1)-\sin(x)}{x\sin(x)}$$

I evaluated ths limit using L'Hôpital's and found $-\frac12$ as the answer.

I eventually ended up to : $$L=\frac12\lim_{x\to 0} 3\cos^2(x)\sin(x)-\frac12$$ I find L'hopital's to be very lengthy in this case.

Is there another way to do it ? I'm stuck

Thanks for the help, T.D.

T.D
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    It's not clear which limit you want to compute: the one in the title of your question or the one in the body? I suppose in the numerator you have $\ln(\tan(x)+1)$, right? – Pantelis Sopasakis Jan 30 '20 at 23:25
  • That's what happens after a copy-paste. Thanks a lot ! – T.D Jan 30 '20 at 23:29
  • Why do you need to not use L'Hospital? One way to do this is by using the Taylor expansion, although I guess if you don't want to use L'Hospital you might not want a solution along these lines either. – YiFan Tey Jan 30 '20 at 23:31
  • Taylor would be fine :) (actually I think L'hopital is a bit of a bruteforce in this case) – T.D Jan 30 '20 at 23:33
  • Taylor polynomials are almost always superior to L'Hôpital's rule (except perhaps in the case where what's involved is just the definition of the derivative and L'Hôpital is redundant to start with). – Ted Shifrin Jan 30 '20 at 23:34
  • So, T.D., what happens when you try to use degree 2 Taylor polynomials for numerator and denominator? – Ted Shifrin Jan 30 '20 at 23:34
  • Well, I am a self-taught maths student, and I am just beginning with Taylor polynomials, thought I think this is a good introductory example then ! – T.D Jan 30 '20 at 23:37

4 Answers4

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Add and subtract $\tan x$ in numerator and then split the expression under limit as $$\frac{\log(1+\tan x) - \tan x} {\tan^2x}\cdot\frac{\tan^2x}{x^2}\cdot\frac{x}{\sin x} +\frac{\tan x - \sin x} {x\sin x} \tag{1}$$ Using L'Hospital's Rule once or via Taylor series one can show that $$\lim_{t\to 0}\frac{\log(1+t)-t}{t^2}=-\frac{1}{2}$$ and putting $t=\tan x$ it is now clear that the first factor in first term in equation $(1)$ tends to $-1/2$ and other factors tend to $1$. Hence the first term in $(1)$ tends to $-1/2$.

The second term in $(1)$ can be rewritten as $$\frac{1-\cos x} {x\cos x} $$ which equals $$\frac{\sin^2x}{x^2}\cdot\frac{x}{\cos x(1+\cos x)} $$ and thus it tends to $0$. It should now be clear that the desired limit is $-1/2$.

L'Hospital's Rule should not be used blindly. Before applying the rule try to simplify the expression using limit laws and standard limits and also try to get expressions where the use of L'Hospital's Rule is simple and efficient. In general if applying L'Hospital's Rule leads to more complicated expressions then you are applying it the wrong way. Also don't use it more than necessary.

Paramanand Singh
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Since $\log(1+x)=x-\frac{x^2}2+O\!\left(x^3\right)$, we have $$ \begin{align} &\frac{\log(1+\tan(x))-\sin(x)}{x\sin(x)}\\ &=\frac{\color{#C00}{\tan(x)}\color{#090}{-\frac{\tan^2(x)}2}+\color{#00F}{O\!\left(\tan^3(x)\right)}\color{#C00}{-\tan(x)\cos(x)}}{x\sin(x)}\\ &=\frac{\color{#C00}{\tan(x)}}{x}\frac{\color{#C00}{1-\cos(x)}}{\sin(x)}\color{#090}{-\frac12}\frac{\color{#090}{\tan^2(x)}}{x\sin(x)}+\frac{\color{#00F}{\tan^2(x)}}{x\sin(x)}\color{#00F}{O(\tan(x))}\\ &=\underbrace{\ \frac{\tan(x)}{x}\vphantom{\frac{()}{1+()}}\ }_1\underbrace{\frac{\sin(x)}{1+\cos(x)}}_0-\frac12\underbrace{\ \frac{\tan(x)}{x\vphantom{()}}\ }_1\underbrace{\ \frac1{\cos(x)}\ }_1+\underbrace{\ \frac{\tan(x)}{x\vphantom{()}}\ }_1\underbrace{\ \frac1{\cos(x)}\ }_1\underbrace{O(\tan(x))\vphantom{\frac1{()}}}_0\\ \end{align} $$

robjohn
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You could expand the transcendental functions in the limitand as their Maclaurin series up to second order: since in a neighborhood of $0$ we have $$\begin{split} \tan x = x + x^2 p(x), \quad \sin x = x + x^2 q(x), \quad \ln(x+1) = x- \frac 1 2 x^2 + x^2 r(x) \end{split}$$ for some functions $p,q,r$ that vanish as $x \to 0$, then $$\begin{split} L &= \lim_{x\to 0} \frac{\ln(x + x^2p(x) + 1)-x - x^2q(x)}{x(x + x^2q(x))} \\ &= \lim_{x\to 0} \frac{1}{1+x q(x)}\frac{(x + x^2 p(x)) - \frac 1 2 (x + x^2p(x))^2 + (x+x^2p(x))^2 r(x)-x- x^2q(x)}{x^2} \\ &= 1 \cdot \lim_{x\to 0}\left( -\frac{1}{2} + p(x)(1-x-x^2p(x)/2)+r(x)(1+2xp(x)+x^2p(x)^2)-q(x) \right)= -\frac 1 2. \end{split}$$ There is a short-hand notation that can help simplifying the above clutter: one writes $$f(x) = g(x) + \mathtt o(h(x)) \quad \text{as}\ x\to 0 $$ if there exists a function $s$ defined in a neighborhood of $0$ such that $f(x) = g(x) + h(x) s(x)$ in that neighborhood. Thus $\tan x = x + \mathtt o(x^2)$, $\sin x = x + \mathtt o(x^2)$ and $\ln(x+1) = x - \frac 1 2 x^2 + \mathtt o(x^2)$. Take a look at this wiki page for further information.

giobrach
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    This is not quite correct: you should expand at order $2$ , so write $+o(x^2)$ in the numerator, and in the denominator expand $x(x+o'(x)$ as $x^2+o(x^2)$. – Bernard Jan 30 '20 at 23:57
  • You can't do it like this man. Don't replace $A$ with $B$ unless $A=B$. Thus $\log(1+x+o(x))$ should equal $$(x+o(x)) - \frac{(x+o(x)) ^2}{2}+o((x+o(x))^2)$$ and $\sin x$ equals $x+o(x^2) $. – Paramanand Singh Jan 31 '20 at 00:06
  • Using Taylor expansions without rigor may give you bad results. – Paramanand Singh Jan 31 '20 at 00:07
  • The way you have written $\sin x =x+o(x) $ it leads to complications while dealing with the term $o(x) /x^2$. – Paramanand Singh Jan 31 '20 at 00:10
  • Better write details as the asker is a novice in this area of Taylor (see his/her comments). The key aspect of learning Taylor series is to learn how to multiply, divide and compose Taylor series upto a given order and even more important is to understand the order till which one needs to develop Taylor series. – Paramanand Singh Jan 31 '20 at 00:12
  • I will correct the hand-wavy $\mathtt o$ in the first step and eviscerate the details in the next edit. I was working under the assumption that OP knew $\tan$ and $\sin$ have no second-order contributions – giobrach Jan 31 '20 at 00:14
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Why not to compose Taylor series from the beginning $$\frac{\log(\tan(x)+1)-\sin(x)}{x\sin(x)}$$ knowing that is must be at least to $O(x^2)$ because of the denominator $$\tan(x)+1=1+x+\frac{x^3}{3}+O\left(x^5\right)$$ $$\log(\tan(x)+1)=x-\frac{x^2}{2}+\frac{2 x^3}{3}-\frac{7 x^4}{12}+O\left(x^5\right)$$ $$\log(\tan(x)+1)-\sin(x)=-\frac{x^2}{2}+\frac{5 x^3}{6}-\frac{7 x^4}{12}+O\left(x^5\right)$$ $$\frac{\log(\tan(x)+1)-\sin(x)}{x\sin(x)}=\frac{-\frac{x^2}{2}+\frac{5 x^3}{6}-\frac{7 x^4}{12}+O\left(x^5\right) } {x^2-\frac{x^4}{6}+O\left(x^5\right) }$$ Now, long division $$\frac{\log(\tan(x)+1)-\sin(x)}{x\sin(x)}=-\frac{1}{2}+\frac{5 x}{6}-\frac{2 x^2}{3}+O\left(x^3\right)$$ which shows the limit and how it is approached.

Moreover, this gives you a shorcut to compute the expression even quite far away from $x=0$. Try it for $x=\frac \pi {24}$ for which we know the exact values of the trigonometric functions (see here) ,the exact value of the expression is $-0.4008$ while the above truncated series would give $-0.4023$.

Claude Leibovici
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  • Great answer :) I have yet to learn how composition works but you definitly helped me – T.D Feb 02 '20 at 21:08