For all x {\displaystyle x} , we have − 1 ≤ sin ( x ) ≤ 1 {\displaystyle -1\leq \sin(x)\leq 1} , and therefore − e x ≤ e x sin ( x ) ≤ e x {\displaystyle -e^{x}\leq e^{x}\sin(x)\leq e^{x}} . Since lim x → − ∞ − e x = lim x → − ∞ e x = 0 , {\displaystyle \lim _{x\rightarrow -\infty }-e^{x}=\lim _{x\rightarrow -\infty }e^{x}=0,} the squeeze theorem then gives lim x → − ∞ e x sin ( x ) = 0 {\displaystyle \color {blue}\lim _{x\to -\infty }e^{x}\sin(x)=0} .
