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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?)

 Feb 8, 2018
 #1
avatar+111331 
+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

 

 

 

cool cool cool

 Feb 8, 2018
edited by CPhill  Feb 8, 2018

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