Question 944970
For the limit to exist, {{{f(x)}}} as you approach from the left ({{{x<1}}}) must equal {{{f(x)}}} as you approach from the right {{{x>=1}}}.
{{{ax^2-5=1/x-2a}}}
{{{a(1)^2-5=1/1-2a}}}
{{{a-5=1-2a}}}
{{{3a=6}}}
{{{a=2}}}
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For it to be continuous then,
{{{c=2}}}
That way the limit from the left, the limit from the right and the value at x=1 are all the same which is the definition of continuous.