Does a Function Have to Be Continuous to Use Direct Substitution on a Limit
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Direct Substitution is a technique that is used to quickly evaluate limits of continuous functions by directly plugging in the value where we need to find the limit. First, we will start talking about continuous and discontinuous functions. Then, we will discuss the direct substitution property in more detail. Finally, we will work through several examples and see how direct substitution works.
What are Continuous Functions?
A function f is continuous at a point p if \displaystyle\lim_{x \to p} f(x)= f(p).
As a result, the above definition implies the following:
- f(p) is defined; in other words, p is in the domain of f and
- \displaystyle\lim_{x \to p} f(x) exists and, moreover,
- \displaystyle\lim_{x \to p} f(x)= f(p).
Continuous functions display no gaps in their graphs. For instance, the following functions are continuous at every point in their domain:
On the other hand, a function f is discontinuous at a point p if \displaystyle\lim_{x \to p} f(x) does not exist. For example, the graphs below are discontinuous. You can see that y= \frac{2}{x- 5} is discontinuous at x= 5 and that y= \frac{1}{x^2} is not continuous at x= 0 .
What is Direct Substitution?
One simple technique that can be applied to evaluating limits in calculus is direct substitution which applies to continuous functions only.
Direct Substitution Property
If f is a continuous function and n is in the domain of f , then \displaystyle\lim_{x \to n} f(x)= f(n).
Please keep in mind that the direct substitution property applies to only to functions which are continuous at a number n ; however, if you are asked to find the limit of a function at a number n where the function is discontinuous, this property does not apply.
To see how this method works, let's take a look at several examples.
Examples
Example 1: Direct Substitution
Find \displaystyle\lim_{x \to -1} {x^3- 5x^2+ 2x- 3}.
Step 1. First, check the domain of the function to see if the function is continuous at the number where we are asked to evaluate the limit. Therefore, we can see that this is a polynomial function which is continuous on all of its domain (the domain of this function is the real numbers), including the value of -1 .
Step2. Second, we have to decide whether to use direct substitution. Therefore, since the function is continuous at -1 , we can use this property to find the limit.
Directly plug in the value of -1 so that
\displaystyle\lim_{x \to -1} {x^3- 5x^2+ 2x- 3} = (-1)^3 -5 (-1)^2+ 2(-1)- 3
Simplify
= -1- 5- 2- 3
Get the final result
= -11.
Example 2: Direct Substitution
Find \displaystyle\lim_{x \to 2}\sqrt {6x- 3}.
Step 1. First, determine the domain of the function. \sqrt {6x- 3} is a polynomial function of degree 0.5 which is continuous on its domain of all real numbers x such that x>= 0.5 . Secondly, we are evaluating the limit as x approaches 2 . Therefore, 2 is in the domain of this function since 2 >= 0.5 .
Step 2. As a result, we can use direct substitution to evaluate the limit.
Plug in the value of 2 so that
\displaystyle\lim_{x \to 2}\sqrt {6x- 3} = \sqrt {6(2)- 3}
Simplify
= \sqrt {12- 3} = \sqrt {9}
Obtain the final answer
= 3 .
Example 3: Direct Substitution
Find \displaystyle\lim_{x \to 0}{\frac{2x^3+7}{x- 3}}.
Step 1. First, figure out what the domain of the function is. {\frac{2x^3+7}{x- 3}} is a rational function which is continuous on its domain of real numbers x such that x \neq 3 (in other words, the denominator of this rational function cannot be 0 ).
Step 2. Therefore, we can directly substitute to find the limit.
Evaluate the limit directly at 0 such that
\displaystyle\lim_{x \to 0}{\frac{2x^3+7}{x- 3}} = \frac{2(0)^3+ 7}{0- 3}
Simplify
= \frac{0+ 7}{-3}
Get the result
= - \frac{7}{3}.
Example 4: Can't use Direct Substitution
Find \displaystyle\lim_{x \to -2} \frac{x^2- 4}{x+ 2}.
Step 1. Check the domain of the function. \frac{x^2 -4}{x+ 2} is continuous on the domain of all real numbers x such that x \neq -2 (to clarify, the denominator cannot be 0 ). However, -2 is exactly the point where the function is discontinuous.
Step 2. As a result, direct substitution doesn't work in this case. We have to apply different laws to calculate this limit.
Practice Problems
Now it's your turn to practice the concepts explained in this tutorial and find the following limits:
1. \displaystyle\lim_{x \to 2} {2x^2+ x - 15}.
2. \displaystyle\lim_{x \to -2}\sqrt {5x^3- 26x- 3}.
3. \displaystyle\lim_{x \to -1} \frac{-3x^2+ 6}{x- 3}.
4. \displaystyle\lim_{x \to -4} \frac{x^2- 16}{x+ 4}.
5. \displaystyle\lim_{x \to 9} \frac{2- x}{\sqrt{x- 5}}.
Practice Solutions
- The solution is -6.
- The answer is 3.
- The solution is \frac{-3}{4}.
- You can't use direct substitution because the function is discontinuous at x= -4 . You need to use a different method to calculate the limit.
- The answer is \frac{-7}{2}.
Source: https://mathleverage.com/direct-substitution/
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