Find Value of K for Continuity of f(x) at x=1

[UrbanXrisis]

let f be a function defined by f(x)=4-x^2 when x=<1 and k+x when x>1
What value of k will f be continuous at x=1?

I know the answer is k=2, however, I don't know how to show to correct work. I got 2 when I sketched a graph of 4-x^2 and plugged in some numbers but I don't know how to show it algebraically/using an actual method.

 

[NateTG]

Are you familiar with epsilon-delta proofs of continuity?

 

[BobG]

To be continous, the limit has to exist (in this case, you have two functions on either side of 1 - they should both have the same limit as they approach 1) and f(x) has to equal the limit.

Find the limit of the first function as x approaches 1 from the left.

[tex]
\lim_{x\rightarrow 1^-} 4-x^2
[/tex]

The limit of the function, f(x)=k+1, as x approaches 1 from the right must equal the limit of the first function.

[tex]
\lim_{x\rightarrow 1^+} k+1 = \lim_{x\rightarrow 1^-} 4-x^2
[/tex]

Substitute the value for k that makes this true.

In other words,

[tex]
\lim_{x\rightarrow 1^+} k+1 = \lim_{x\rightarrow 1^+} k+\lim_{x\rightarrow 1^+}1
[/tex]
k and 1 are both constants, so the limit of k = k and the limit of 1 = 1.
[tex]
\lim_{x\rightarrow 1^+} k+\lim_{x\rightarrow 1^+}1 = k+1 = 3
[/tex]

so, subtracting 1 from both sides, k=2