Answer:a
Step-by-step explanation:
Given
[tex]f(x)=1.6^x\sin 3x[/tex]
we know that value of sin function vary between [tex]-1\ \text{to}\ 1[/tex]
i.e. it take finite value so
[tex]\lim \limits_{x\rightarrow -\infty}f(x)=\lim \limits_{x\rightarrow -\infty}\quad 1.6^x\sin 3x[/tex]
when [tex]x\rightarrow -\infty[/tex]
then [tex]1.6^x\rightarrow 0[/tex]
therefore
[tex]\lim \limits_{x\rightarrow -\infty}\quad 1.6^x\sin 3x=0[/tex]
option a is correct
