Q. 78 - Calculus

Question

Prove that for all k > 100, the quantity $\frac{1}{k^2}$ is in the interval (0, 0.0001). What does this have to do with the limit of the sequence

$\left\{\frac{\mathrm{1}}{{\mathrm{k}}^{\mathrm{2}}}\right\} \mathrm{as} k\to \infty ?$

Step-by-Step Solution

Answer

The given statement is proved.

Step 1. Given Information.

The given limit of the sequence is $\left\{\frac{\mathrm{1}}{{\mathrm{k}}^{\mathrm{2}}}\right\} \mathrm{as} k\to \infty .$

Step 2. Proving.

Let the function is $f\left(k\right)=\frac{1}{k^2}.$

Take the limit of the above function as $k\to {100}^{+}.$

$$\begin{gathered} \underset{k\to {100}^{+}}{\lim }f\left(k\right)=\underset{k\to {100}^{+}}{\lim }\frac{1}{k^2} \\ \underset{k\to {100}^{+}}{\lim }f\left(k\right)=\frac{1}{{100}^2} \\ \underset{k\to {100}^{+}}{\lim }f\left(k\right)=\frac{1}{10000} \\ \underset{k\to {100}^{+}}{\lim }f\left(k\right)=0.0001 \end{gathered}$$

Thus, for all $k>100,$ the quantity $\frac{1}{k^2}$ is in the interval $\left(0,0.0001\right).$