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Use the definition of continuity and the properties of limits to show that the function is continuous at the given number $ a $.
$ g(t) = \frac{t^2 + 5t}{2t + 1}, \hspace{5mm} a = 2 $
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04:06
Daniel Jaimes
Calculus 1 / AB
Chapter 2
Limits and Derivatives
Section 5
Continuity
Limits
Derivatives
Oregon State University
Baylor University
University of Michigan - Ann Arbor
Lectures
04:40
In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.
In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.
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for this problem, we need to show that the function G ft, she is equal to t squared plus five T over to T plus one. It's continuous at equal to two. By definition, a function is continuous at a point. If the following conditions hold, the first one would be pour the function to be defined at the given point. That means that the function exists at this point. And then you have limit as X approaches A of F of X exists. That means The one sided limits are equal. So limit as extra purchased A from the left. Other function equals limit as X approaches A from the right of this function. And then lastly, we need to show that the limit as X approaches A ah F of X. This is equal to F. Of A. So using this definition we have GF two which is equal to two squared plus five times 2 Over two times 2 Plus one. This gives us a value 14/5. And so the function exists at two. For the second part, we need to show that the limit S. T approaches to of G f T. This is equal to limit as T approaches to uh T squared plus five T Over two T plus one exist. Now, evaluating at two. We get two squared plus five times 2 Over two times 2 Plus one. This gives the value of 14/5 and have shown that the limit exists at two. Since the limit and the value of the function are equal. Then you have shown The 3rd part, which states that the limit as T approaches to Of G. of Thi, this is equal to GF two. Therefore GFT is continuous at two and its limit is equal to 14/5.
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