Limit and Maclaurin series

Question

I'm trying to solve this limit.

$\lim_{x\rightarrow0}\frac{(e^x \sin x- (x+1)\tan x)}{(x\log\cos(x))}$

But my result, using Maclaurin series, is:$-\frac{2}{3}$ And it should be: $0$

Answer 1

Expanding to the 3rd order

• $x\log\cos(x)=-\frac{x^3}2+ o(x^3)$
• $e^x \sin x=(1+x+x^2/2+o(x^2))(x-x^3/6+o(x^3))=x+x^2+x^3/3+o(x^3)$
• $(x+1)\tan x=(x+1)(x+x^3/3+o(x^3))=x+x^2+x^3/3+o(x^3)$

thus

$$\lim_{x\rightarrow0}\frac{(e^x \sin x- (x+1)\tan x)}{(x\log\cos(x))}=\lim_{x\rightarrow0}\frac{o(x^3)}{\frac{x^3}2+ o(x^3)}=\lim_{x\rightarrow0}\frac{o(1))}{\frac{1}2+ o(1)}=0$$

Answer 2

after three times applying L'Hospital we get $$-2\,{\frac {{{\rm e}^{x}} \left( \cos \left( x \right) \right) ^{5}-{ {\rm e}^{x}}\sin \left( x \right) \left( \cos \left( x \right) \right) ^{4}+2\,x \left( \cos \left( x \right) \right) ^{2}+2\, \left( \cos \left( x \right) \right) ^{2}-3\,\cos \left( x \right) \sin \left( x \right) -3\,x-3}{ \left( 2\,x\sin \left( x \right) +3\, \cos \left( x \right) \right) \cos \left( x \right) }} $$ plugging $x=0$ in this term we get $0$

Answer 3

We can take $\sin x$ as a factor from numerator and since $(\sin x) /x\to 1$ the desired limit is equal to the limit of the expression $$\frac{e^x-(x+1)\sec x} {\log\cos x} $$ which is same as that of (using the fact that $\lim_{t\to 1}\dfrac{\log t} {t-1}=1$) $$\frac{e^x-x-1+(x+1)(1-\sec x)} {\cos x-1}$$ and this splits into two terms as $$\frac{e^x-1-x}{x^2}\cdot\frac{x^2}{\cos x-1}+\frac{x+1}{\cos x} $$ The limit is clearly equal to $(1/2)(-2)+1=0$.