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02 Limits and Continuity

Limits

Let \(f\) be defined @ all points in some neighborhood of a point \(x_0\)

Then \(L = \lim\limits_{x \to x_0} f(x)\) is limit for \(f(x)\) when \(x \to x_0\) if for a given \(\epsilon > 0\), there exists a \(\delta > 0\) such that \(|x-x_0| < \delta \implies |f(x)-L| < \epsilon\)

Finding \(\delta\)

  1. Solve the inequality \(f(x) - L < \epsilon\) for \(x\)
  2. Find an interval \((a, b)\) such that \(a \le x_0 \le b\)
  3. Choose \(\delta = \min (x_0-a, b - x_0)\)

This choice places the interval \((x_0 - \delta, x_0 + \delta)\) within \((a, b)\)

One-sided Limits

Let \(f\) be defined at all points in the neigborhood of \(x_0\) (in particular to right of \(x_0\)), then \(f\) is said to have the right-hand limit \(L\), when \(x\) approaches \(x_0\) from the right if the following conditions are satisfied:

For a given \(\epsilon > 0\), there exists a \(\delta > 0\) such that

  • \(x_0 < x < x_0 + \delta\)
  • \(|f(x) - L| < \epsilon\)

The limit is represented as

\[ L = \lim_{x \to {x_0}^+} f(x) = f({x_0}^+) \]

Similarly, we define the left-hand limit

While working on one-sided problms, we proceed as follows

\[ \begin{aligned} f({x_0}^+) &= \lim_{h \to 0} f(x_0 + h), & h > 0 \\ f({x_0}^-) &= \lim_{h \to 0} f(x_0 - h), & h > 0 \end{aligned} \]

Continuity

A function \(f(x)\) is continuous @ a point \(x_0\) if the following conditions are satisfied

  1. \(f(x_0)\) exists
  2. \(\lim_{x \to x_0} f(x)\) (Both LHL and RHL) exists
  3. \(\lim_{x \to x_0} f(x) = f(x_0)\)

Note

If \(f\) and \(g\) are continuous functions in a domain \(D\), then the following functions are also continuous in all points of F

\[ \begin{aligned} f \pm g \\ fg \\ \frac f g \\ kf, & (k \text{= const}) \end{aligned} \]

The following functions are known to be continuous in their domain of definition

  1. polynomial
  2. exponential
  3. trignometric
Last Updated: 2023-01-25 ; Contributors: AhmedThahir

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