Definition of Limit of a Function
Solved Problems
Example 3.
Using the \(\varepsilon-\delta-\) definition of limit, find the number \(\delta\) that corresponds to the \(\varepsilon\) given with the following limit: \[\lim\limits_{x \to 7} \sqrt {x + 2} = 3,\,\varepsilon = 0.2\]
Solution.
According to the definition of limit,
Substituting \({f\left( x \right)}\) and \(\varepsilon\), we get
Square all three parts of the inequality.
which is equivalent to the inequality
Thus, we can choose the number \(\delta = 1.16\) to satisfy the original inequality for \(\varepsilon.\)
Example 4.
Prove that \[\lim\limits_{x \to \infty } {\frac{{x + 1}}{x}} = 1.\]
Solution.
We can apply the similar techniques to limits at \(x \to \infty .\) Assume \(\varepsilon \gt 0.\) We must have
We can choose a number \(N,\) depending on \(\varepsilon,\) such that \(\left| x \right| \gt N.\) Then the inequality \(\left| {{\frac{1}{x}}} \right| \lt \varepsilon \) will be satisfied if
This means that at large \(N\) (as \(x \to \infty \))
The graph of the function is sketched in Figure \(1.\)
Example 5.
Prove that \[\lim\limits_{x \to \infty } {\frac{{2x - 3}}{{x + 1}}} = 2.\]
Solution.
We assume that \(\varepsilon \gt 0.\) Find a number \(N\) such that for any \(x \gt N\) the following inequality is valid:
Convert this inequality to get
Since \(0 \lt x \lt N,\) then \(x + 1 \gt 0,\;\) so we can simply write
By setting \(N = {\frac{5}{\varepsilon }} - 1\) (or \(N = 0\) if this difference is negative), we obtain
This means that (see Figure \(2\))
