Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
p(v)=2 sqrt(3v^(2)+1), a=1

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Best Answer

f(x) is continuous at x=a if and only if
lim_(x -> a)f(x)=f(a)
lim_(v -> 1)p(v)= lim_(v -> 1)2 sqrt(3v^(2)+1)
Using the constant law for limits, we can write
=2* lim_(v -> 1) sqrt(3v^(2)+1)
Use theorem 9 to simplify lim_(x -> 2) sqrt[3](x^(2)+4)
=2 sqrt(lim_(v -> 1)3v^(2)+1)
Using the sum law for limits, we can write
=2 sqrt(lim_(v -> 1)3v^(2)+ lim_(v -> 1)1)
Recall that: lim_(x -> a)x^(n)=a^(n)
=2 sqrt(3*1^(2)+1)=p(1)
Hence proved that p(x) is continuous at x=1

**Result:**Hint: we have to show that lim_(v -> 1)p(v)=p(1)