Question

consider the following values for x:
x
-0.1
-0.01
-0.001
0.001
0.01
0.1
if (f(x)=\frac{1}{x^{3}}), evaluate f(x) for each value and determine if (lim_{x
ightarrow0}f(x)) exists/ is infinite.
the limit does not exist
the limit exists
the limit is positive infinity
the limit is negative infinity

Explanation:

Step1: Evaluate for negative x - values

When \(x=-0.1\), \(f(-0.1)=\frac{1}{(-0.1)^3}=\frac{1}{- 0.001}=-1000\). When \(x = - 0.01\), \(f(-0.01)=\frac{1}{(-0.01)^3}=\frac{1}{-0.000001}=-1000000\). When \(x=-0.001\), \(f(-0.001)=\frac{1}{(-0.001)^3}=\frac{1}{-0.000000001}=- 1000000000\). As \(x\) approaches \(0\) from the left - hand side (\(x\to0^{-}\)), \(f(x)\to-\infty\).

Step2: Evaluate for positive x - values

When \(x = 0.001\), \(f(0.001)=\frac{1}{(0.001)^3}=\frac{1}{0.000000001}=1000000000\). When \(x = 0.01\), \(f(0.01)=\frac{1}{(0.01)^3}=\frac{1}{0.000001}=1000000\). When \(x = 0.1\), \(f(0.1)=\frac{1}{(0.1)^3}=\frac{1}{0.001}=1000\). As \(x\) approaches \(0\) from the right - hand side (\(x\to0^{+}\)), \(f(x)\to+\infty\).

Step3: Determine the limit

Since \(\lim_{x\to0^{-}}f(x)=-\infty\) and \(\lim_{x\to0^{+}}f(x)=+\infty\), the two - sided limit \(\lim_{x\to0}f(x)\) does not exist.

Answer:

The limit does not exist