Question

(a) How large do we have to take \(x\) so that \(1/{x^2} < 0.0001\)?

(b) Taking \(r = 2\)in Theorem 5, we have the statement

\(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{1}{{{x^2}}} = 0\)

Prove this directly using Definition 7.

Short answer

(a) If \(x > 100\)then \(\frac{1}{{{x^2}}} < 0.0001\)

(b) It is proved that \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{1}{{{x^2}}} = 0\).

Step-by-step solution

(a) Step 1: Simplify the given condition

Perform cross product then take square root to simplify.

\(\begin{array}{c}\frac{1}{{{x^2}}} < 0.0001\\{x^2} > \frac{1}{{0.0001}}\\{x^2} > 10000\\x > 100\end{array}\).

Thus, \(x > 100\) so that \(\frac{1}{{{x^2}}} < 0.0001\).

(b) Step 2: Precise Definition of a limit at infinity  

Assume \(f\) be any function defined on the interval \(\left( {a,\infty } \right)\). Then \(\mathop {{\rm{lim}}}\limits_{x \to \infty } f\left( x \right) = L\) ,

which implies that for every \( \in > 0\) there exist a number \(N\) such that if \(x > N\) then \(\left| {f\left( x \right) - L} \right| < \in \).

Given \( \in > 0\), we have to find \(N\) such that if \(x > N\) then \(\left| {\frac{1}{{{x^2}}} - 0} \right| < \in \).

Simplify the inequality

Simplify the inequality \(\frac{1}{{{x^2}}} < \in \) for \(x\).

\(\begin{array}{c}\frac{1}{{{x^2}}} < \in \\{x^2} > \frac{1}{ \in }\\x > \frac{1}{{\sqrt \in }}\end{array}\)

Choose \(N = \frac{1}{{\sqrt \in }}\) which satisfy if \(x > \frac{1}{{\sqrt \in }}\) then \(\left| {\frac{1}{{{x^2}}} - 0} \right| < \in \).

By using Precise Definition of a limit at infinity, \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{1}{{{x^2}}} = 0\).

Hence, it is proved that \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{1}{{{x^2}}} = 0\).