Question

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{x-4}{\left|x-4\right|}+a,\mathrm{if}& x<4\\ a+b,\mathrm{if}& x=4\\ \frac{x-4}{\left|x-4\right|}+b,\mathrm{if}& x>4\end{array}\right.$is continuous at x = 4, find a, b.

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{c}\frac{x-4}{\left|x-4\right|}+a,\mathrm{if}\mathrm{x}<4\\ a+b,\mathrm{if}\mathrm{x}=4\\ \frac{x-4}{\left|x-4\right|}+b,\mathrm{if}\mathrm{x}>4\end{array}\right.$

We observe
(LHL at x = 4) = $\underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(4-h\right)$
$=\underset{h\to 0}{\lim }\left(\frac{4-h-4}{\left|4-h-4\right|}+a\right)=\underset{h\to 0}{\lim }\left(\frac{-h}{\left|-h\right|}+a\right)=a-1$


(RHL at x = 4) = $\underset{x\to 4^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(4+h\right)$
$=\underset{h\to 0}{\lim }\left(\frac{4+h-4}{\left|4+h-4\right|}+b\right)=\underset{h\to 0}{\lim }\left(\frac{h}{\left|h\right|}+b\right)=b+1$

And $f\left(4\right)=a+b$

If f(x) is continuous at x = 4, then

​$\underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{x\to 4^{+}}{\lim }f\left(x\right)=f\left(4\right)$
$\Rightarrow a-1=b+1=a+b$
$\Rightarrow a-1=a+b,b+1=a+b$
$\Rightarrow b=-1,a=1$