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Patience Owens

Open question

2022-08-19

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{3 v^{2} + 1}, a = 1

glasinamaav

Beginner2022-08-20Added 10 answers

f(x) is continuous at x=a if and only if

lim_{x \Rightarrow a} f (x) = f (a)

lim_{v \Rightarrow 1} p (v) = lim_{v \Rightarrow 1} 2 \sqrt{3 v^{2} + 1}

Using the constant law for limits, we can write

= 2 \cdot lim_{v \Rightarrow 1} \sqrt{3 v^{2} + 1}

Use theorem 9 to simplify

lim_{x \Rightarrow 2} \sqrt{3} {x^{2} + 4}

= 2 \sqrt{lim_{v \Rightarrow 1} 3 v^{2} + 1}

Using the sum law for limits, we can write

= 2 \sqrt{lim_{v \Rightarrow 1} 3 v^{2} + lim_{v \Rightarrow 1} 1}

Recall that:

lim_{x \Rightarrow a} x^{n} = a^{n}

= 2 \sqrt{3 \cdot 1^{2} + 1} = p (1)

Hence proved that p(x) is continuous at x=1
Result:
Hint: we have to show that

lim_{v \Rightarrow 1} p (v) = p (1)

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