The function f:RR defined by [f(x)=_n(2 x)-x^2n(x-1)1+x^2n] is continuous for all x in:

Question

The function $f:\mathbb{R}\to\mathbb{R}$ defined by
$$f(x)=\lim_{n\to\infty}\frac{\cos(2\pi x)-x^{2n}\sin(x-1)}{1+x^{2n}}$$
is continuous for all $x$ in:

Accepted Answer

For $|x|<1$, $x^{2n}	o0$, so $f(x)=\cos(2\pi x)$ --- continuous.

For $|x|>1$, dividing numerator and denominator by $x^{2n}$: $f(x)=\dfrac{-\sin(x-1)}{x-1}$ --- continuous.

At $x=1$: $f(1)=\dfrac{\cos2\pi-\sin0}{2}=\dfrac12$. But $\lim_{x	o1^-}\cos(2\pi x)=1$ and $\lim_{x	o1^+}\dfrac{-\sin(x-1)}{x-1}=-1$. These don't match $f(1)$ or each other, so $f$ is discontinuous at $x=1$.

At $x=-1$: similarly $f(-1)$ from the definition doesn't match the two one-sided limits, so $f$ is discontinuous at $x=-1$.

Hence $f$ is continuous for all $x\in\mathbb{R}-\{-1,1\}$.

Solutions · 1