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Problem 12 Easy Difficulty
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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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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