Answer
$y=\frac{\sqrt{x^4+1}}{1+\sin^2x}$ is continuous on $(-\infty,\infty)$
Work Step by Step
$$y=\frac{\sqrt{x^4+1}}{1+\sin^2x}$$
- Domain: $y$ is defined where $x^4+1\ge0$ and $1+\sin^2x\ne0$:
For $x^4+1\ge0$, since $x^4\ge0$, so $x^4+1\gt0$ for all $x\in R$
For $1+\sin^2x\ne0$, since $\sin^2x\ge0$, so $\sin^2x+1\gt0$ for all $x\in R$
So our domain here is $(-\infty,\infty)$
1) Examine $y=\sqrt{x^4+1}$
We know that $\lim_{x\to c}(x^4+1)=c^4+1$ on $(-\infty,\infty)$
So $y=x^4+1$ is continuous on $(-\infty,\infty)$
Applying Theorem 8, the root of a function continuous at $x=c$ is also continuous at $x=c$.
Therefore, $y=\sqrt{x^4+1}$ is continous on $(-\infty,\infty)$.
2) Examine $y=1+\sin^2x$
We know that $\lim_{x\to c}(1+\sin^2x)=1+\sin^2c$ on $(-\infty,\infty)$
So $y=1+\sin^2x$ is continuous on $(-\infty,\infty)$
3) According to Theorem 8, the quotient of two functions continuous at $x=c$ is also continuous at $x=c$ (as long as the denominator does not equal $0$).
Therefore, $y=\frac{\sqrt{x^4+1}}{1+\sin^2x}$ is continuous on $(-\infty,\infty)$
