consider the following values for x:\nx\n-0.1\n-0.01\n-0.001\n0.001\n0.01\n0.1\nif (f(x)=\frac{1}{x^{3}})…
consider the following values for x:\nx\n-0.1\n-0.01\n-0.001\n0.001\n0.01\n0.1\nif (f(x)=\frac{1}{x^{3}}), evaluate f(x) for each value and determine if (lim_{x\rightarrow0}f(x)) exists/ is infinite.\nthe limit does not exist\nthe limit exists\nthe limit is positive infinity\nthe limit is negative infinity
Answer
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