MA005: Calculus I

The following definition states precisely what is meant by the phrase "we can guarantee that the values of $\mathrm{f}(\mathrm{x})$ are arbitrarily close to $\mathrm{K}$ by using sufficiently large values of $\mathrm{x}$".

Definition: $\lim _{x \rightarrow \infty} \mathrm{f}(\mathrm{x})=\mathrm{K}$ means for every given $\varepsilon > 0$, there is a number $N$ so that if $x$ is larger than $\mathrm{N}$ then $f(x)$ is within $\varepsilon$ units of $K$ (equivalently; $|f(x)-K| < \varepsilon$ whenever $x > N$.)

Example 10: Show that $\lim \limits_{x \rightarrow \infty} \frac{x}{2 x+1}=\frac{1}{2}$.

Solution: Typically we need to do two things. First we need to find a value of $\mathrm{N}$, usually depending on $\varepsilon$. Then we need to show that the value of $\mathrm{N}$ we found satisfies the conditions of the definition.

(i) Assume that $|\mathrm{f}(\mathrm{x})-\mathrm{K}|$ is less than $\varepsilon$ and solve for $\mathrm{x} > 0$.
If $\varepsilon > \left|\frac{x}{2 x+1}-\frac{1}{2}\right|=\left|\frac{2 x-(2 x+1)}{2(2 x+1)}\right |= | \frac{-1}{4 x+2} |=\frac{1}{4 x+2}$, then $4 x+2 > \frac{1}{2}$ and

$x > \frac{1}{4}\left(\frac{1}{\varepsilon}-2\right)$. For any $\varepsilon > 0$, take $\mathrm{N}=\frac{1}{4}\left(\frac{1}{\varepsilon}-2\right)$.

(ii) For any $\varepsilon > 0$, take $\mathrm{N}=\frac{1}{4}\left(\frac{1}{\varepsilon}-2\right)$. (Now we can just reverse the order of the steps in part (i). ) If $x > 0$ and $x > N=\frac{1}{4}\left(\frac{1}{\varepsilon}-2\right)$, then $4 x+2 > \frac{1}{2}$ so $\varepsilon > \frac{1}{4 x+2}=\left|\frac{x}{2 x+1}-\frac{1}{2}\right|=|f(x)-K|$.
We have shown that "for every given $\varepsilon$, there is an $\mathrm{N}$ " that satisfies the definition.