Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (b)/Solution 1

For all $x$, we have $-1\le \sin (x)\le 1$, and therefore $-e^x\le e^x\sin (x)\le e^x$. Since $\underset{x\to -\mathrm{\infty }}{\lim }-e^x=\underset{x\to -\mathrm{\infty }}{\lim }{e}^x=0,$ the squeeze theorem then gives $\underset{x\to -\mathrm{\infty }}{\lim }{e}^x\sin (x)=0$.