# If g(x)=x if x<0, x^2 if #0<= x <=1#, x^3 if x>1, how do you show that g is continuous on (all real #'s)?

##### 1 Answer

The function

# g(x) = { (x,x<0), (x^2, 0 le x le 1), (x^3, x > 1) :} #

Which we can graph as follows:

The individual functions

**When #x=0#**

Consider the left-hand limit:

# lim_(x rarr 0^-) g(x) = lim_(x rarr 0) x \ \= 0 #

And the right-hand limit

# lim_(x rarr 0^+) g(x) = lim_(x rarr 0) x^2 = 0 #

And the value of the function:

# g(0) = 0 #

So

**When #x=1#**

Consider the left-hand limit:

# lim_(x rarr 1^-) g(x) = lim_(x rarr 1) x^2 \ \= 1 #

And the right-hand limit

# lim_(x rarr 1^+) g(x) = lim_(x rarr 1) x^3 = 1 #

And the value of the function:

# g(1) = 1 #

So

Hence