Question 1102893
{{{f(x)=5x-11}}},{{{x<1}}}
{{{f(x)=1}}},{{{x=1}}}
{{{f(x)=3x+6}}},{{{x>1}}}
As x approaches 1 from the left,
{{{f(x)->5(1)-11}}}
{{{f(x)->-6}}}
As x approaches 1 from the right,
{{{f(x)->3(1)+6}}}
{{{f(x)->9}}}
Since the limit approaching from the left and right are not equal, in addition the value at {{{x=1}}} doesn't equal either value, then the limit does not exist.