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.
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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.
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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.
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Since $\log(1+x)=x-\frac{x^2}2+O\!\left(x^3\right)$, we have $$ \begin{align} &\frac{\log(1+\tan(x))-\sin(x)}{x\sin(x)}\\ &=\frac{\color{#C00}{\tan(x)}\color{#090}{-\frac{\tan^2(x)}2}+\color{#00F}{O\!\left(\tan^3(x)\right)}\color{#C00}{-\tan(x)\cos(x)}}{x\sin(x)}\\ &=\frac{\color{#C00}{\tan(x)}}{x}\frac{\color{#C00}{1-\cos(x)}}{\sin(x)}\color{#090}{-\frac12}\frac{\color{#090}{\tan^2(x)}}{x\sin(x)}+\frac{\color{#00F}{\tan^2(x)}}{x\sin(x)}\color{#00F}{O(\tan(x))}\\ &=\underbrace{\ \frac{\tan(x)}{x}\vphantom{\frac{()}{1+()}}\ }_1\underbrace{\frac{\sin(x)}{1+\cos(x)}}_0-\frac12\underbrace{\ \frac{\tan(x)}{x\vphantom{()}}\ }_1\underbrace{\ \frac1{\cos(x)}\ }_1+\underbrace{\ \frac{\tan(x)}{x\vphantom{()}}\ }_1\underbrace{\ \frac1{\cos(x)}\ }_1\underbrace{O(\tan(x))\vphantom{\frac1{()}}}_0\\ \end{align} $$
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 $-0.4023$.