Question

which of the following functions have the property: $lim_{x
ightarrow -infty} f(x)=0$? (choose all that apply.)
$f(x)=x^{3}+sqrt{x^{6}+3x^{2}}$
$f(x)=\frac{1}{x}$
$f(x)=\frac{x - 1}{x^{2}+9}$
$f(x)=e^{x}$
$f(x)=\frac{x^{2}}{x}$
$f(x)=ln(-x)$

Explanation:

Step1: Analyze $f(x)=x^{3}+\sqrt{x^{6}+3x^{2}}$

As $x\to -\infty$, $\sqrt{x^{6}+3x^{2}}=\vert x^{3}\vert\sqrt{1 + \frac{3}{x^{4}}}\approx - x^{3}$ (since $x<0$), then $x^{3}+\sqrt{x^{6}+3x^{2}}\to x^{3}-x^{3}=0$ as $x\to -\infty$.

Step2: Analyze $f(x)=\frac{1}{x}$

By the limit rule $\lim_{x\to-\infty}\frac{1}{x}=0$.

Step3: Analyze $f(x)=\frac{x - 1}{x^{2}+9}$

Divide numerator and denominator by $x^{2}$: $\lim_{x\to-\infty}\frac{\frac{x}{x^{2}}-\frac{1}{x^{2}}}{1+\frac{9}{x^{2}}}=\lim_{x\to-\infty}\frac{\frac{1}{x}-\frac{1}{x^{2}}}{1+\frac{9}{x^{2}}}=0$.

Step4: Analyze $f(x)=e^{x}$

As $x\to -\infty$, $\lim_{x\to-\infty}e^{x}=0$ because the exponential - function $y = e^{x}$ decays to 0 as $x$ approaches negative infinity.

Step5: Analyze $f(x)=\frac{x^{2}}{x}=x$

$\lim_{x\to-\infty}x=-\infty$.

Step6: Analyze $f(x)=\ln(-x)$

As $x\to -\infty$, $\ln(-x)\to+\infty$.

Answer:

$f(x)=x^{3}+\sqrt{x^{6}+3x^{2}}$, $f(x)=\frac{1}{x}$, $f(x)=\frac{x - 1}{x^{2}+9}$, $f(x)=e^{x}$