Notice that both pieces of [tex]f(x)[/tex] are continuous on their respective domains. That is, [tex]e^{6x}[/tex] exists and is defined for all [tex]x\le0[/tex]; the same goes for [tex]9-8\cos x[/tex] for all [tex]x>0[/tex].
a. Because of the continuity of each piece, we have
[tex]\lim_{x\to-1}f(x)=f(-1)=e^6[/tex]
where we take [tex]f(x)=e^{6x}[/tex] because that's how [tex]f[/tex] is defined if [tex]x\le0[/tex].
b. For [tex]f[/tex] to be continuous at [tex]x=0[/tex], both one-sided limits have to come out to be the same number, [tex]f(0)[/tex]. By definition of
The limits are
[tex]\displaystyle\lim_{x\to0^-}f(x)=\lim_{x\to0}e^{6x}=1[/tex]
[tex]\displaystyle\lim_{x\to0^+}f(x)=\lim_{x\to0^+}(9-8\cos x)=9-8=1[/tex]
c. [tex]x\to-\infty[/tex] means [tex]x[/tex] is negative, for which [tex]f(x)=e^{6x}[/tex]. Then
[tex]\displaystyle\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}e^{6x}=0[/tex]
