Limit Find the value of $c$ given function
Find $c$, such that the function
$$f(x) = \begin{cases}
& {1 – \sqrt{x} \over x – 1},&0\le x < 1\\
& c,& x= 1
\end{cases}$$
is continuous for all $x \in [0, 1 ]$.
I try to solve that question on interval $x = 0$ i get answer $-1$, with $f(0)$, and get undefined on limit so this is discontinous.
but when the $x$ is $1$ how can i solve? and How i get value of $c$?
i try the following approach
on left hand limit we get $- 1 / 2 $ but on right hand limit we have nothing!. to me this function seems to not complete. this was the assignment i told our teacher to upload solution of this, but he didnt do that till yet.
Sorry, if it is nonsense question; i am software engineer student and on 2nd sem we're studying calculus subject.
Best Answer
(Rewriting this answer)
That's fine. The requirement for continuity $$\lim\limits_{x\to x_0}f(x) = f(x_0)$$ implicitly means that as $x\to x_0$, it does so through values of $x\neq x_0$ that are in the domain of $f$.
So if $x_0 = 1$, all of the domain values of $f$ are the to left of $x_0$. You are therefore implicitly computing a one-sided limit, and $x\nearrow 1$.
The limit depends only on values of $f(x)$ for $x<1$. If this happens to equal the value of $f(x)$ for $x=1$, then $f$ is continuous at $x=1$.