Suppose $f$ is defined on $\mathbb R$ and $$\lim_{x\to 1}f(x)=5.$$ I need to prove that $$\lim_{x\to 0}f(\sin(x)+1)= 5$$ using only the $\delta$-$\epsilon$ definitions of limits and continuity.
I have proved that $\sin{x}+1$ is continuous at $0$, which means $\lim_{x\to0}\sin{x}+1=\sin{0}+1=1$. This gives me $$ \forall\eta>0, \exists\delta>0\text{ such that }0<|x-0|<\delta\implies\left|\sin{x}+1-1\right|<\eta. \tag{1}\label{eq1} $$ From $\lim_{y\to0}f(y) = 5$ I know that $$ \forall\epsilon>0, \exists\eta>0\text{ such that }0<|y-1|<\eta\implies\left|f(y)-5\right|<\epsilon. $$ Take $y=\sin{x}+1$ then $$ \forall\epsilon>0, \exists\eta>0\text{ such that }0<|\sin{x}+1-1|<\eta\implies\left|f(\sin{x})-5\right|<\epsilon. \tag{2}\label{eq2} $$ I want to combine equations \eqref{eq1} and \eqref{eq2}, but then I need to prove that for the chosen $\delta$ neighbourhood of $x$, $|\sin{x}+1-1|\neq0$, except at $x=0$.
I can't go any further from here.
Am I on the right track? Are there any errors in the rough work I have so far? How can I move on from this stage?