# In the limit lim_(t to oo) 1/(1+4t)=0 , how do you find B>0 such that whenever t>B, 1/(1+4t)<0.01?

Jul 30, 2017

Just solve the inequality $\frac{1}{1 + 4 t} < 0.01$ to get $t > 24.75$ (so $B = 24.75$).

#### Explanation:

The inequality $\frac{1}{1 + 4 t} < 0.01 = \frac{1}{100}$ is equivalent to $1 + 4 t > 100$ (flip the fractions and reverse the direction of the inequality). This, in turn, is equivalent to $4 t > 99$ and equivalent to $t > \frac{99}{4} = 24.75$.

For a function $f$ defined for all $t > 0$, the definition of ${\lim}_{t \to \infty} f \left(t\right) = L$ can be stated like this: for all $\epsilon > 0$, there exists a number $B$ such that $| f \left(t\right) - L | < \epsilon$ for all $t > B$. (The value of $f \left(t\right)$ can be made to be a distance from $L$ less than any $\epsilon > 0$ if the input $t$ is sufficiently large...$B$ is the measure of "sufficiently large").

The point of this problem for $f \left(t\right) = \frac{1}{1 + 4 t}$ and $L = 0$ is to find the corresponding value of $B$ when $\epsilon = 0.01$. This does not prove that ${\lim}_{t \to \infty} f \left(t\right) = 0$, but is equivalent to the kind of algebra you would need to do to help you do a proof.

Assuming $t > 0$, we have $\frac{1}{1 + 4 t} < \epsilon \setminus \leftrightarrow 1 + 4 t > \frac{1}{\epsilon} \setminus \leftrightarrow t > \frac{\frac{1}{\epsilon} - 1}{4} = \frac{1 - \epsilon}{4 \epsilon}$, we can do the proof as follows:

Proof: Let $\epsilon > 0$ be given and let $B = \frac{1 - \epsilon}{4 \epsilon}$. Suppose $t > 0$ also satisfies $t > B$. It follows that $1 + 4 t > \frac{1}{\epsilon}$ and therefore $0 < f \left(t\right) = \frac{1}{1 + 4 t} < \epsilon$ (the first inequality is obvious here since $t > 0$). Therefore, $| f \left(t\right) - 0 | < \epsilon$. This proves that ${\lim}_{t \to \infty} f \left(t\right) = 0$.