Find the following limits or state that they do not exist. Ass...

Question

Find the following limits or state that they do not exist. Assume a, b , c, and k are fixed real numbers.

lim t→5 (1/t^2 − 4t − 5 −1/ 6(t − 5))

Accepted Answer

Welcome back, everyone calculate the limit below or state that it does not exist limit of one divided by X squared minus six X minus seven minus one divided by eight, multiplied by X minus seven. When X approaches seven, there are four answer choices. A says 1/7 B negative 1 32nd C negative 164 in the, the limit does not exist. So basically, first of all, we can apply direct substitution limit when X approaches seven of one divided by X squared minus six X minus seven minus one divided by eight X minus seven. If we apply direct substitution, well, we get one divided by zero, right? 49 minus seven minus 42. This gives us zero. So one divided by zero gives us infinity. And then we are subtracting one divided by zero, which is infinity. And this is an indeterminate form. Now, since it is an indeterminate form, we want to somehow eliminate it and see if this limit exists. So what we're going to do is base lap life factorization and then find the common denominator with everything onto one fraction. Then see what we get. So let's rewrite the limit X approaches seven. Let's factorize the first denominator and essential. We can use the AC method which gives us X squared minus six, X minus seven is equal to X minus seven multiplied by X plus one. So we get two roots, one of them is seven, the other one is negative one. And this essentially gives us our factored expression. We are subtracting one divided by eight X minus seven. And now our goal is to use the common denominator. So for the first fraction, we only need to include eight. So we're multiplying both sides by eight. For the second fraction, we need X plus one. So we're multiplying both sides by X plus one. Let's rewrite the resultant limit limit. When X approaches seven. Now we can use the common denominator eight multiplied by X minus seven, multiplied by X plus one. For the first fraction we get eight and we are subtracting X plus one. Now lets simplify limit when X approaches 78 minus X minus one, this essentially simplifies to seven minus X. In the denominator, we have eight X minus seven X plus one. Now seven minus XX minus seven. Those are quite similar, right? And we can actually get that X minus seven in the numerator if we add a negative sign and then we can flip the signs X minus seven. If we distribute the negative sign, we get seven minus X, right, which is basically the same expression as we had before. So that's a valid expression. The reason why it's useful is because now we can clearly see that we can cancel out one of the factors X minus seven. So now our limit becomes negative limit because we had the negative sign, we can factor out that negative sign negative limit. When X approaches seven of one divided by eight X plus one, let's apply the direct substitution. Once again, we get negative one divided by eight in parentheses, seven plus one. So we end up with negative one divided by eight squared, eight multiplied by eight. That's 64. So now we have a finite number and this means that the limit is negative 1 64th which is answer choice. C Thank you for watching.

Find the following limits or state that they do not exist. Assume a, b , c, and k are fixed real numbers.

lim t→5 (1/t^2 − 4t − 5 −1/ 6(t − 5))

Key Concepts

Limits

Indeterminate Forms

L'Hôpital's Rule

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