Question

Let $f\left(x\right)=\{\begin{array}{cccc}a{\sin }^{2n}x& forx\ge 0& & n\to \infty \\ b{\cos }^{2m}x-1& forx<0& & m\to \infty \end{array}$ where $a$ and $b$ are finite real values

• A. $f\left(0^{-}\right)\ne f\left(0^{+}\right)$
• B. $f\left(0^{+}\right)\ne f\left(0\right)$
• C. $f\left(0^{-}\right)\ne f\left(0\right)$
• D. $f$ is continuous at $x=0$

Answer 1

Answer: (A), (C)

$f\left(0^{-}\right)\ne f\left(0\right)$
$f\left(0^{-}\right)\ne f\left(0^{+}\right)$

$f\left(0^{+}\right)=a.(sinx{)}^{2n}=0$
$f\left(0\right)=0andf\left(0^{-}\right)=-1$
$f\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{n\to \infty }{lim}a.(sinx{)}^{2n}=a(0^{+}{)}^{\infty }=0$
$f\left(0^{-}\right)=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{m\to \infty }{lim}b.(cosx{)}^{2m}-1=b(1^{-}{)}^{\infty }-1$
$x=-1$
and $f\left(0\right)=0$
Hence, option 'A' is correct.