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# Evaluate the limit, if it exists. $\displaystyle \lim_{t \to -3}\frac{t^2 - 9}{2t^2 + 7t + 3}$

## $\frac{6}{5}$

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So here we are given example of a certain limit. We have the limit as t approaches negative three of t square in minus nine divided by two, T squared plus 70 plus three. So let's see if we can factor the numerator and denominator. So the numerator factories too t minus three Times T- Plus three. And we can see that the denominator factors to to T plus one times T plus three. And the first thing we can see here is that uh we don't simplify this. We would have an indeterminant form since this would be equivalent to negative 21 plus three. Mhm plus 18. So this would be zero and the numerator is also zero. So 0/0 is an indeterminate form. So without this indeterminant form we can basically with this indeterminant form we have to find ways of simplifying. So we can cross out the T plus threes. And now we can see we don't have any determinant form anymore since we don't evaluate to zero divided by zero. And we can just use direct substitution and we find that this is equivalent to negative six divided by negative five. And so this is this limit is equivalent to 6/5. We could have also evaluated this biological rule, although we would need two iterations of lobster rule since we have to uh evaluate the numerator and denominator twice in this case. And besides that, this is our final answer

Johns Hopkins University

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