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.
Best Answer
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.