If $f(x)=\left\{\begin{matrix}\sin\frac{πx}{2}&;x<1\\4x-3&;1≤x≤2\\\log_2(2x^2-4)&;2<x<3\end{matrix}\right.$ then value of $\underset{x→1}{\lim}f(x)+\underset{x→2^+}{\lim}f(x)$ is:
Answer & explanation
Correct answer: option 2
At x = 1: $\underset{at\,x=1}{L.H.L}=\underset{h→0}{\lim}f(1-h)=\underset{h→0}{\lim}\sin\frac{π}{2}(1-h)=\underset{h→0}{\lim}\cos\frac{π}{2}h=1$
$\underset{at\,x=1}{R.H.L}=\underset{h→0}{\lim}f(1+h)=\underset{h→0}{\lim}4(1+h)-3=1$
As L.H.L. = R.H.L., limit exists at x = 1
At x = 2: $\underset{at\,x=2}{L.H.L}=\underset{h→0}{\lim}f(2-h)=\underset{h→0}{\lim}4(2-h)-3=5$
$\underset{at\,x=2}{R.H.L}=\underset{h→0}{\lim}f(2+h)=\underset{h→0}{\lim}\log_2(2(2+h)^2-4)=\log_2\,4=2$
$\underset{x→1}{\lim}f(x)+\underset{x→2^+}{\lim}f(x)$
$=1+2=3$