A not straightforward limit The original limit is:
$$\lim_{x \to +\infty} \frac{e^{\sin \frac{1}{x}}-1-\frac{1}{x}}{\ln \left(1+\frac{x^2}{(1+x)^3} \right )-\frac{x^2}{(1+x)^3}}$$
I performed the substitution $y=\frac{1}{x}$ which leads to
$$\lim_{y \to 0^{+}} \frac{e^{\sin y}-1-y}{\ln \left ( 1+\frac{y}{(1+y)^3} \right )-\frac{y}{(1+y)^3 }}$$
now using the Maclaurin polynomials it should lead to the more simple limit
$$\displaystyle \lim_{ y\to 0^{+}} \left ( -\frac{1}{6} \frac{y + o(y^3)}{\frac{1}{(1+y)^6}+o\left (\frac{y}{(1+y)^3}\right )} \right )  $$
which is zero.
Could anyone tell me if this solution is right please?
 A: That is not what I have.
$\lim_\limits{y\to 0} \frac {e^{\sin y} -1 - y}{\ln (1+ f(y)) - f(y)}$
The Talor expansion of the numerator is
$1 + \sin y + \frac 12 \sin ^2 y+\cdots - 1 - y$
Expanding each of those sine functions
$1+ y - \frac 16y^3 +\cdots + \frac 12 y^2 - \frac 16 y^4+\cdots -1-y$
Which will ultimately be
$\frac 12 y^2 + o(y^3)$
For the denominator, the Taylor expanson of $\ln (1+f(y)) = f(y) - \frac 12 (f(y))^2 + \cdots$
giving us $\frac {\frac 12 y^2 + o(y^3)}{-\frac 12 \frac {y^2}{(1+y)^6} + o(y^3)}$
As the limit approaches $0$ we will get $-1$
A: Let $t=y/(1+y^3).$ Divide the denominator and the numerator by $t^2.$ The denominator takes the form $${\log (1+t)-t\over t^2}\underset{t\to 0}{\rightarrow} -{1\over 2}$$ by applying l'Hospital one time or by  $\log(1+t)=t-t^2/ 2 +O(t^3).$ The numerator divided by $y^2$ (as $(1+y)^6$ tends to $1$) takes the form
$${e^{\sin y}-1-y\over y^2}={e^{\sin y}-1-\sin y+O(y^3)\over y^2}\approx {1\over 2}{\sin^2y\over y^2}\approx {1\over 2}\quad y\to 0$$ Hence the limit is equal $-1.$
Remark The main difference between this and the other answers is $$e^{\sin y}-1-y=e^{\sin y}-1-\sin y+O(y^3)$$ which avoids  $$e^{\sin y}-1=\sin y+{1\over 2}\sin^2y+O(y^3)$$
