Question

Let $[.]$ denote the greatest integer function.
Consider $f\left(x\right)=\{\begin{array}{cc}2-\left|x\right|,& -1\le x\le 1\\ |x-2|-x,& 1<x\le 3\end{array}$ and $g\left(x\right)=\{\begin{array}{cc}\sin x-1,& 0\le x<\frac{\pi }{2}\\ \left[x\right]-\cos (x-2),& \frac{\pi }{2}\le x\le \pi \end{array}.$

Which of the following statements is/are CORRECT?

• A. $\underset{x\to 1^{+}}{lim}g\left(f\right(x\left)\right)=-1$
• B. $\underset{x\to \pi /2^{-}}{lim}g\left(f\right(g\left(x\right)\left)\right)=0$
• C. $\underset{x\to 2^{+}}{lim}\frac{f\left(g\right(x\left)\right)}{f\left(x\right)-2}=\frac{1}{2}$
• D. $\underset{x\to 0^{+}}{lim}\frac{g\left(f\right(x\left)\right)}{\left(f\right(x)-2{)}^2}=\frac{1}{2}$

Answer 1

Answer: (B), (D)

$\underset{x\to 0^{+}}{lim}\frac{g\left(f\right(x\left)\right)}{\left(f\right(x)-2{)}^2}=\frac{1}{2}$

$\underset{x\to 1^{+}}{lim}g\left(f\right(x\left)\right)=\underset{x\to 0^{-}}{lim}g\left(x\right)$ which does not exist.

$\underset{x\to \pi /2^{-}}{lim}g\left(f\right(g\left(x\right)\left)\right)$
$=\underset{x\to 0^{+}}{lim}g\left(f\left(g(\frac{\pi }{2}-x)\right)\right)=\underset{x\to 0^{+}}{lim}g\left(f(\cos x-1)\right)$
$=\underset{x\to 0^{+}}{lim}g(2-|\cos x-1|)=0$

$\underset{x\to 2^{+}}{lim}\frac{f\left(g\right(x\left)\right)}{f\left(x\right)-2}=\underset{x\to 0^{+}}{lim}\frac{f\left(g\right(2+x\left)\right)}{f(2+x)-2}=\underset{x\to 0^{+}}{lim}\frac{f(2-\cos x)}{-4}$
$=\underset{x\to 0^{+}}{lim}\frac{|2-\cos x-2|-(2-\cos x)}{-4}=\underset{x\to 0^{+}}{lim}\frac{2\cos x-2}{4}=0$


$\underset{x\to 0^{+}}{lim}\frac{g\left(f\right(x\left)\right)}{\left(f\right(x)-2{)}^2}=\underset{x\to 0^{+}}{lim}\frac{g(2-x)}{(2-x-2{)}^2}$
$=\underset{x\to 0^{+}}{lim}\frac{[2-x]-\cos (2-x-2)}{x^2}$
$=\underset{x\to 0^{+}}{lim}\frac{1-\cos x}{x^2}=\frac{1}{2}$