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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 $

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

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This is problem number twelve of the Stuart Calculus Safe Edition Section two point five. Use the definition of continuity and the properties of limits to show that the function is continuous at the given number, eh? G of T is equal to the quantity. T squared was five divided by the quantity to t plus one for ese quoted too. The definition of continuity is shown here. The limiters experts is a private function. I must be equal to the elements Existing vehicle to the function evaluated at the value of a Ah, we first began with evaluating the Lim Ah first, So we take the limit of dysfunction. So in this case, half is gonna be G. It's x X approaches A which is too Ah, dysfunction here. Ah, the quantity two squared plus fifteen over to depose one. Once a publication that we can do is use our limit lines and use a limerick of division Ah in order for us to separate this into two limits as expressions, too of the numerator t squared plus five team and the limit his expertise to of the denominator to t plus one. And this is possible because the denominator. You know you needed a T equals two. Um, just a clarification. Each of these limits is as t approaches to yeah, and not x approaches to since we're using team. But we know that a teacher approaches to it is our approach to zero. And so that is the only conditions in order for us to use this quotient law on. And now each of the terms that happened about him have ah sum that we can separate this limit one limited to two limits. So it becomes trending well down on work. Down here, filament. Is it team purchase too? T squared. Plus, and then we're gonna use a constant long as well to remove the firing from the limit he purchased two of team all the ready by ah two times a limit as team purchase too. Team. Plus the limited's T approaches to off one. And now we cannot evaluate each of these limits. The power law here tells us that this is equal to to the second power plus five times the limit. Is he purchased? Two tea is equal to to terminus the limit as t approaches to of tea which is too less limited to purchase to have one and the limit of any constant value. Uh, no matter what he approaches is kind of a constant value. And this will evaluate this. We should get four plus ten minutes fourteen and in the denominator four plus one, which is five. And we found a limit on the left side. Ah, and then we want to evaluate what today is. We should see that when we play gin A, we will be doing the exact same Ah, calculation. We get the same answer fourteen over five because we're directly plugging in two into every tea in this function. And as we can tell, it's thie exact same work. So we'LL get fourteen or five and thus we will have proven that dysfunction is continuous. Harry equals to two.

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