How do you explain why $f(x) = x^2$ if $x>=1$ and 2x if x<1 is continuous at x = 1?

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Answer 1 · 2016-11-30T11:01:32

It is not continuous at $1$ , because the limit at $1$ does not exist.

$f(x) = {(x^2,"if",x >=1),(2x,"if",x < 1) :}$

$f$ is continuous at $1$ if and only if

$lim_(xrarr1)f(x) = f(1)$ . $" "$ (existence of both is implied)

We see from the definition of $f$ above that

$f(1) = (1)^2 = 1$ .

In finding the limit, we'll need to consider the left and right limits separately because the rule changes at $x=1$ .

The limit from the right is

$lim_(xrarr1^+)f(x) = (1)^2 = 1$ .

The limit from the left is

$lim_(xrarr1^-)f(x) = 2(1) = 2$ .

Because the right and left limits are not the same, the limit does not exist.

Therefore the function is not continuous at $1$ .

How do you explain why f(x) = x^2f(x) = x^2 if x>=1x>=1 and 2x if x<1 is continuous at x = 1?