Answer to Question #122135 in Calculus for moreen

Question

Consider the piecewise function
F(x) = - x+1, IF x <1
X-1, IF 1<x<2
5-xsquared, IF x>equalto 2
(i) Find lim x arrow 1f(x) if it exists.
(ii) Show that f (x) is continuous at x=2.
(iii) Sketch the graph of f(x)

Answer 1

Consider the piecewise function

$f(x)=\begin{cases} -x+1, x<1; \\ x-1, 1<x<2; \\ 5-x^2, x\ge 2.\end{cases}$

(i) As we are looking for the limit of a piecewise defined function at the point where the function changes its formula, then we have to take one-sided limits separately since different formulas will apply depending on which side we are approaching from.

$\lim_{x\to1-} f(x)=\lim_{x\to1-} (-x+1)=-1+1=0.$

$\lim_{x\to1+} f(x)=\lim_{x\to1+} (x-1)=1-1=0.$

Since both limits give 0, $\lim_{x\to1} f(x)=0.$

(ii) Notice that $f(2)=5-2^2=1.$

We need to look at the one-side limits at $x=2.$

$\lim_{x\to2-} f(x)=\lim_{x\to2-} (x-1)=2-1=1.$

$\lim_{x\to2+} f(x)=\lim_{x\to2+} (5-x^2)=5-2^2=1.$

$\lim_{x\to2-} f(x)=\lim_{x\to2+} f(x)=f(2)$ hence $f$ is continuous at x=2.

(iii) The graph of $f(x):$