Question

(a) How large do we have to take \(x\) so that \(1/\sqrt x < 0.0001\)?

(b) Taking \(r = \frac{1}{2}\) in Theorem 5, we have the statement

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

Prove this directly using Definition 7.

Short answer

(a) If \(x > {10^8}\) then \(\frac{1}{{\sqrt x }} < 0.0001\).

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

Step-by-step solution

(a) Step 1: Simplify the given condition  

Perform cross product then take square to simplify.

\(\begin{array}{c}\frac{1}{{\sqrt x }} < 0.0001\\\sqrt x > \frac{1}{{0.0001}}\\\sqrt x > 10000\\x > {\left( {10000} \right)^2}\\x > {10^8}\end{array}\).

Thus, \(x > {10^8}\) so that \(\frac{1}{{\sqrt x }} < 0.0001\).

(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 \(\left| {\frac{1}{{\sqrt x }} - 0} \right| < \in \).

Simplify the inequality

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

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

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

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

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