Show that \(f(x) = x\sqrt{x+3}\) is continuous at the point x = -1 using the definition of continuity. (It must show the three conditions of continuity satisfied)

So far I've boiled it down to:

f(-1) = ??

\(\lim_{x\rightarrow -1} f(x)\) = ??

\(\lim_{x\rightarrow -1} f(x) = f(-1)\) (How to prove?)

CurlyFry Feb 8, 2018

#1**+1 **

1. If the function is continuous at x = -1, f(-1) and lim as x → -1 from both sides must have the same value

f(-1) = -1 √ [ - 1 + 3 ] = - √2

And

lim x √ [ x + 3 ] = (-1) √ [-1 + 3] = -√2

x → -1

So this is true

2. Also f(a) must = the limit as it approaches "a" from the right and the left

These are going to look similar

lim x √ [ x + 3 ] = (-1) √ [-1 + 3] = -√2 = f(-1)

x → -1^{+}

And

lim x √ [ x + 3 ] = (-1) √ [-1 + 3] = -√2 = f(-1)

x → -1^{-}

So...we have proved that

1. The limit - from both sides - is the same as the function value at x = -1

2. The limit from the right side is the same as the function value at x = -1

3. The limit from the left side is the same as the function value at x = -1

CPhill Feb 8, 2018