Limits & Continuity
Piecewise limit; continuity of function on ℝ
MJMT_Full_Test_11
Grade 12
Question:
Let $f:\mathbb{R}\to\mathbb{R}$ be defined as $f(x)=\displaystyle\lim_{p\to\infty}\frac{\sin\!\left((2x-1)\frac{\pi}{2}\right)\!x^{4p}e^{x^2-1}+x^{4p}}{1+x^{4p+2}-x^{4p}}$. $f(x)$ is continuous for all $x$ in
$\mathbb{R}$
$\mathbb{R}-\{1\}$
$\mathbb{R}-\{-1\}$
$\mathbb{R}-\{-1,1\}$
Step-by-Step Solution
Key Concept: Analyse by cases: $|x|<1$: $x^{4p}\to0$, $f(x)=0$. $|x|>1$: divide by $x^{4p}$, $f(x)=[\sin((2x-1)\pi/2)\cdot e^{x^2-1}+1]/(x^2-1)$. $x=\pm1$: compute directly.
$f$ is discontinuous at $x=\pm1$; continuous on $\mathbb{R}-\{-1,1\}$.
Correct Answer: 4