1/17/2024 0 Comments Extended continuity calculusAnswerįor a function □ ( □ ) to be continuous at a we need three conditions to hold: Given □ ( □ ) = □ + □ − 2 □ − 1 , if possible or necessary, define □ ( 1 ) so that □ is continuous at □ = 1. Let’s consider an example where we use the definition of continuity at a point to determine how we can extend a function to be continuous at a value outside of its domain.Įxample 2: Using Continuity to Find the Value Needed to Extend a Function’s Domain Hence, the second continuity condition of □ at □ = □ 2 also holds.įinally, since this is also equal to □ □ 2 , we can conclude that the third continuity condition also holds and so the function □ is continuous at □ = □ 2. Therefore, both the left and right limits exist and are equal to − 7, so we have shown We can then evaluate this limit by direct substitution: We can do the same for the right limit where we note that when □ > □ 2, we have that □ ( □ ) = 6 2 □ − 1 c o s, giving us Since this is a trigonometric expression, we can evaluate this limit by direct substitution: We start with the left limit and note that when □ ≤ □ 2, we have □ ( □ ) = − 7 □ + 7 □ s i n c o s, giving us To check the second condition for continuity, we will check whether the left and right limits of □ at □ = □ 2 both exist and are equal. So, the first condition for continuity at □ = □ 2 holds. Therefore, □ 2 is in the domain of □ and □ □ 2 = − 7. In our case □ = □ 2, we can see from the definition of □ that l i m → □ ( □ ) and □ ( □ ) must have the same value.s i n c o s c o s Answerįor a function □ ( □ ) to be continuous at □, we need three conditions to hold: In our first example, we will determine the continuity of a piecewise-defined function at the endpoints of its subdomains.Įxample 1: Discussing the Continuity of a Piecewise-Defined Function Involving Trigonometric Ratios at a Pointĭiscuss the continuity of the function □ at □ = □ 2, given Since all three conditions hold, we have shown that □ ( □ ) = | □ | is continuous at 0. Third, we have found the values of □ ( 0 ) and l i m → □ ( □ ) and shown that both of these are equal to the same value, 0. Therefore, the left and right limits of | □ | at 0 are equal, so Similarly, for the right limit, the values of □ are all positive, so First, when evaluating l i m → | □ |, the values of □ are all negative, so | □ | = − □ in this limit. We can use this to evaluate the left and right limits. We need to recall the piecewise definition of the modulus function: To evaluate this limit, we recall that we can determine whether a limit exists by checking if the left and right limits at this point both exist and are equal. Second, we need to determine l i m → | □ |. l i m → □ ( □ )and □ ( □ ) must have the same value.įor example, let’s check the continuity of □ ( □ ) = | □ | at □ = □.įirst, we know that □ ( 0 ) = | 0 | = 0, so □ = 0 is in the domain of □.(This is equivalent to saying both the left and right limits of □ ( □ ) at □ = □ exist and are equal.) □ must be defined at □ ( □ is in the domain of □).To check if the function □ ( □ ) is continuous at □ = □, we need to check whether the following three conditions hold. How To: Checking Whether a Function Is Continuous at a Point
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