suppose (g(x)) is a continuous function on ((-infty,infty)). which of the following is false?\nthe function…

Explanation:

Step1: Analyze the first option

We know that $\lim_{x\rightarrow0}\frac{\sin(x)}{x} = 1$. Since $g(x)$ is continuous on $(-\infty,\infty)$, and $9 + |g(x)|>0$ for all $x$. The function $\frac{\sin(x)}{9 + |g(x)|}$ is continuous at $x = 0$ as $\lim_{x\rightarrow0}\frac{\sin(x)}{9+|g(x)|}=\frac{\lim_{x\rightarrow0}\sin(x)}{9 + |g(0)|}=0$.

Step2: Analyze the second option

We have $\left|\frac{\sin(x)\cos(x)}{|x^{3}|+|g(x)|}\right|\leq\frac{1}{|x^{3}|+|g(x)|}\leq\frac{1}{|x^{3}|}$. Since $\lim_{x\rightarrow-\infty}\frac{1}{|x^{3}|}=0$, by the Squeeze - Theorem, $\lim_{x\rightarrow-\infty}\frac{\sin(x)\cos(x)}{|x^{3}|+|g(x)|}=0$. So the limit exists.

Step3: Analyze the third option

Since $\lim_{x\rightarrow0}(e^{x}-1)=0$ and $g(x)$ is continuous at $x = 0$, then $\lim_{x\rightarrow0}(e^{x}-1)g(x)=\lim_{x\rightarrow0}(e^{x}-1)\cdot\lim_{x\rightarrow0}g(x)=0\cdot g(0)=0$.

Step4: Analyze the fourth option

Since $g(x)$ is continuous on $(-\infty,\infty)$, it is continuous at $x = 1$. By the definition of continuity, $\lim_{x\rightarrow1}g(x)=g(1)$, so the limit exists.

Step5: Analyze the fifth option

Since $g(x)$ is continuous at $x = 0$, $\lim_{x\rightarrow0}g(x)=g(0)$. Also, the function $y = \sqrt{|u|}$ is continuous at $u = g(0)$. Then $\lim_{x\rightarrow0}\sqrt{|g(x)|}=\sqrt{|g(0)|}$, so the limit exists.

Answer:

The limit $\lim_{x\rightarrow-\infty}\frac{\sin(x)\cos(x)}{|x^{3}|+|g(x)|}$ does NOT exist. (This option is false)