Mathematics - Continuity - Differentiability Question with Solution | TestHub

We evaluate $f(x)$ for different ranges of $x$: If $|x|<1$, $x^{2n}\to 0$ and $x^{2n-1}\to 0$ as $n\to \infty$. So $f(x)=a{x}^2+bx$. If $|x|>1$, $f(x)={\lim }_{n\to \infty }\frac{x^{-1}+a{x}^{2-2n}+b{x}^{1-2n}}{1+x^{-2n}}=\frac{x^{-1}}{1}=\frac{1}{x}$. If $x=1$, $f(1)=\frac{1+a+b}{2}$. If $x=-1$, $f(-1)=\frac{-1+a-b}{2}$. For $f(x)$ to be continuous at $x=1$: ${\lim }_{x\to 1^{-}}f(x)={\lim }_{x\to 1^{+}}f(x)=f(1)$.$a+b=1=\frac{1+a+b}{2}$. From this, we get $a+b=1$. For $f(x)$ to be continuous at $x=-1$: ${\lim }_{x\to -1^{-}}f(x)={\lim }_{x\to -1^{+}}f(x)=f(-1)$.$-1=a-b=\frac{a-b-1}{2}$. From this, we get $a-b=-1$.

Solving the system of equations $a+b=1$ and $a-b=-1$, we add them to get $2a=0⟹a=0$. Subtracting them gives $2b=2⟹b=1$. Therefore, $a+b=0+1=1$.