How to Determine Continuity of Lnxx29
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The concept of continuity is closely related to evaluating limits in calculus. In this tutorial, we will talk about the continuity of a function at a point. Then, we will discuss the different types of discontinuities that occur. Finally, we will illustrate the concepts with several examples.
Definition of Continuity at a Point
A function f is continuous at a point a if \displaystyle\lim_{x \to a}{f(x)}= f(a).
The definition above implies the following if f is continuous at a point a :
- f(a) is defined; in other words, a is in the domain of f , and
- \displaystyle\lim_{x \to a}{f(x)} exists, and, finally,
- \displaystyle\lim_{x \to a}{f(x)}= f(a).
A function is continuous at a specific point a if there are no gaps or breaks and the graph can be drawn without removing pen from paper.
Furthermore, if a function is continuous at a point a , this implies that it is continuous from the right and from the left at a point a .
We know that a function f is continuous from the right at a point a if \displaystyle\lim_{x \to a^+}{f(x)}= f(a) .
We know that a function f is continuous from the left at a point a if \displaystyle\lim_{x \to a^-}{f(x)}= f(a) .
Definition of Discontinuity at a Point
Conversely, if any of the three conditions (implied by the definition of a continuous function at a point a ) breaks, then we say that the function is discontinuous at point a .
When this happens, the discontinuity can be classified into three categories:
- Removable Discontinuity– this means that the discontinuity can be removed by redefining the function f at point a ; in other words, \displaystyle\lim_{x \to a}{f(x)} exists which means that \displaystyle\lim_{x \to a}{f(x)}= N, where N is a real number and \displaystyle\lim_{x \to a}{f(x)} \neq f(a).
- Infinite Discontinuity– where the discontinuity is present at a vertical asymptote; this means that \displaystyle\lim_{x \to a^-}{f(x)}= \pm \infty or \displaystyle\lim_{x \to a^+}{f(x)}= \pm \infty.
- Jump Discontinuity– where the function f has a gap at point a ; in other words, \displaystyle\lim_{x \to a^-}{f(x)} and \displaystyle\lim_{x \to a^+}{f(x)} both exist (and they are not \pm \infty ). However, \displaystyle\lim_{x \to a^-}{f(x)} \neq \displaystyle\lim_{x \to a^+}{f(x)} so that the \displaystyle\lim_{x \to a}{f(x)} does not exist.
Here's what the discontinuities look like:
How to Determine Continuity at a Point
Problem Solving Strategy
1. Determine if f(a) exists.
- If f(a) does not exist, the function is discontinuous at a .
- If f(a) is defined, then go to step 2.
2. Find \displaystyle\lim_{x \to a}{f(x)} . Depending on the function, we may need to determine \displaystyle\lim_{x \to a^-}{f(x)} and \displaystyle\lim_{x \to a^+}{f(x)} .
- If \displaystyle\lim_{x \to a^-}{f(x)}= \displaystyle\lim_{x \to a^+}{f(x)} , then \displaystyle\lim_{x \to a}{f(x)} exists and we can move on to step 3.
- Otherwise, f is discontinuous at a .
3. Check to see if f(a)= \displaystyle\lim_{x \to a}{f(x)} .
- If this happens, the function f is continuous at a .
- Otherwise, it is discontinuous at a .
Examples
Let's now look at a few examples to illustrate continuity at a point and the removable, infinite and jump discontinuities.
Example 1: Finding Continuity at a Point
Find \displaystyle\lim_{x \to -1}{x^2- 5x+ 2}.
We can take a look at the graph of this function.
This is a polynomial function of degree 2 which is continuous on the domain of \mathbb{R} . -1 is in the domain of the function which means that f is continuous at the point -1 . We can use the limit laws or Direct Substitution to find the limit:
\displaystyle\lim_{x \to -1}{x^2- 5x+ 2}= (-1)^2- 5(-1)+ 2= 1+ 5+ 2= 8.
Example 2: Finding Continuity at a Point
Calculate \displaystyle\lim_{x \to 2}{\frac{x^2- 5x+ 6}{x- 2}} and specify the discontinuity type, if applicable.
Let's a take a look at this function by plotting it.
First of all, this is a rational function which is continuous on the domain of \mathbb{R} such that x \neq 2 . However, we are asked to evaluate the limit at x= 2 , precisely where the function is not defined. f(2) is not defined for this function and f is discontinuous at x= 2 .
Nevertheless, we can still calculate \displaystyle\lim_{x \to 2}{\frac{x^2- 5x+ 6}{x- 2}} through algebraic manipulation.
\displaystyle\lim_{x \to 2}{\frac{x^2- 5x+ 6}{x- 2}}= \displaystyle\lim_{x \to 2}{\frac{(x-2)(x- 3)}{x- 2}} = \displaystyle\lim_{x \to 2} (x-3)
We can now use Direct Substitution since f is now continuous at x= 2 .
= 2-3 = -1
To summarize, f is discontinuous at x= 2 because f(2) is not defined, but \displaystyle\lim_{x \to 2}{\frac{x^2- 5x+ 6}{x- 2}}= -1 . Therefore, x= 2 is a removable discontinuity. Check out how you can calculate this in Mathematica using the Mathematica Limit.
Example 3: Finding Continuity at a Point
Find \displaystyle\lim_{x \to 0}{f(x)} where f(x)= \begin{cases} \frac{2}{x^3}, \text { if } x \neq 0 \\ 1, \text { if } x= 0 \end{cases} and specify the type of discontinuity, if applicable.
We can also take a look at this function by plotting its graph below:
First of all, we see if we can calculate f at x= 0 . f(0)= 1 , according to this piecewise function.
Secondly,
\displaystyle\lim_{x \to 0^-}{f(x)}= \displaystyle\lim_{x \to 0^-}{\frac{2}{x^3}}= -\infty
and
\displaystyle\lim_{x \to 0^+}{f(x)}= \displaystyle\lim_{x \to 0^+}{\frac{2}{x^3}}= +\infty
so that \displaystyle\lim_{x \to 0}{f(x)} does not exist. Therefore, the function f has an infinite discontinuity at x= 0 .
Example 4: Finding Continuity at a Point
Determin \displaystyle\lim_{x \to 1}{f(x)} where f(x)= \begin{cases} -x^2+ 1, \text { if } x<= 1 \\ 4x- 5, \text { if } x> 1 \end{cases} and specify the type of discontinuity, if applicable.
Here is what the function looks like:
We can see that f(x) is a rational function which is defined for all values of x on the domain of \mathbb{R} . Also, f(1)= 0 .
Furthermore,
\displaystyle\lim_{x \to 1^-}{f(x)}= \displaystyle\lim_{x \to 1^-}{(-x^2+ 1)}= -1^2+ 1= 0.
and
\displaystyle\lim_{x \to 1^+}{f(x)}= \displaystyle\lim_{x \to 1^+}{(4x- 5)}= 4- 5= -1.
Therefore, \displaystyle\lim_{x \to 1}{f(x)} does not exist and f has a jump discontinuity at x= 1 .
Practice Problems
Now it's your turn to apply the concepts you learned in this tutorial. Determine the following limits and classify the types of discontinuities, if available:
- \displaystyle\lim_{x \to -5}{\frac{1}{x+ 2}}
- \displaystyle\lim_{x \to 1}{\frac{2x+ 5}{x- 1}}
- Find the discontinuities of \displaystyle\lim_{x \to 6}{(x+ \sqrt{x- 5})}
- \displaystyle\lim_{x \to 1}{f(x)} where f(x)= \begin{cases} 2x^2, \text { if } x \neq 1 \\ 4, \text { if } x= 1 \end{cases}
- \displaystyle\lim_{x \to 3}{f(x)} where f(x)= \begin{cases} -x^2+ 5, \text { if } x<= 3 \\ 4x- 3, \text { if } x> 3 \end{cases}
Practice Solutions
- f(x) is continuous at x= -5 because f(-5) exists and \displaystyle\lim_{x \to -5}{\frac{1}{x+ 2}}= -\frac{1}{3}.
- f(1) does not exist and the function is discontinuous at x= 1 ; \displaystyle\lim_{x \to 1}{\frac{2x+ 5}{x- 1}} does not exist and x= 1 is an infinite discontinuity.
- f(x) is continuous at x= 6 and \displaystyle\lim_{x \to 6}{(x+ \sqrt{x- 5})}= 7.
- f(x) has a removable discontinuity at x= 1 .
- Jump discontinuity at x= 3 .
Source: https://mathleverage.com/continuity-at-a-point/
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