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Question 1. Do value of k and m exist such that the following limit exists?Sx2 + kx-2m Ilm x2 + mx-4If so, find the values of k, m and the corresponding limit; If n...

Question

Question 1. Do value of k and m exist such that the following limit exists?Sx2 + kx-2m Ilm x2 + mx-4If so, find the values of k, m and the corresponding limit; If not; explain why not;

Question 1. Do value of k and m exist such that the following limit exists? Sx2 + kx-2m Ilm x2 + mx-4 If so, find the values of k, m and the corresponding limit; If not; explain why not;



Answers

Does a value of $k$ exist such that the following limit exists?
$$\lim _{x \rightarrow 2} \frac{3 x^{2}+k x-2}{x^{2}-3 x+2}$$
If so, find the value of $k$ and the corresponding limit. If not,
explain why not.

So we need to find the values of K that make this limit exists. And the the limit as X approaches negative too, of X squared plus four X plus K over X plus two. So the numerator is in the form A plus B times A plus C. So we need it to be squares. So this is, let's just say the limit as X approaches negative too. So the numerator is in the form A squared plus B plus C, times A plus B C. So we have B plus C is equal to four. B times C is equal to K. So that makes okay equal to yeah, the time. See, So if B is equal to two since X plus two in the denominator we get that K Is equal to four. So this would split two, X plus two. You would need the X plus two to cancel X plus four over, exposed to. So K is equal to four.

So we need to find the value of K. So that the limit will exist. Currently. If you plug in infinity you will get infinity over infinity, which is an indeterminant form. So what the method is for these problems when you're going to infinity is you are factoring out the highest power in the denominator and that needs to cancel one of the factors in the numerator. So this would be the limit. As X goes to infinity. We would need to factor out an X squared. So in fact our next square, this would be one plus three X. The negative. One plus five X to the negative too. Because remember one over infinity is approximately equal to zero. In order for that to happen, K must be at least two. Um Kay is greater than or equal to two.

For the given problem, we want to find the value of the constant K. Such that the limit will exist. Yeah, we're gonna have X squared Plus four. Act plus K. And then we're gonna divide that by exposed to. Um And then we recognize that this we ultimately want to have X plus two. On the top hand X plus some other value. Um And then that is going to all be divided by X-plus two. That way we can cancel this out. I ended up getting what we're looking for. Well since this is going to be right here, We would divide that by two. So the two. Um That way we know that this is going to be an X plus two squared. So that means our K. Value must equal for and that's going to be our final answer.

So we need to find the limit as x approaches one of x squared minus K, X plus four over x minus one. So we need to find where what value of K makes this limit exists. So right now, if we plug in X As one, you'll get an indeterminant form. So we need to do some factoring. So if we factor the numerator, this is the limit As X Approaches one, that means this numerator will become X was the times X plus C Over X -1, and B is -1. So we have X -1 if we need this to be true. And so we also have that BCB times C is equal to four. The since B is negative one, C musical the negative four negative K is equal to B plus C. Since we factored, Which means that K needs to be equal to five.


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