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→2+ |2t − 4|t^2 − 4

Accepted Answer

Welcome back, everyone determine the following limit or state that it does not exist limit of the absolute value of five Z minus 25 divided by Z squared minus 25. When Z approaches five from the right, there are four answer choices. A says 1/4 B, one third, C one half and D, the limit does not exist since we have the absolute value in our limit. Well, essentially we first will have to convert it into one of the parts of the original piece wise function. Let's recall that the absolute value of F of X is basically piecewise function, right? And we know that it can either be F of X itself or negative F of X depending on what F of X is. If F of X is a positive value, then we can just drop the absolute value sign, right. And the absolute value of F of X is equal to F of X if F of X is greater than or equal to zero, otherwise it's negative F of X if F of X is negative. So in this case, if we are approaching five from the right, we can essentially substitute Z equals 55, multiplied by five, gives us 2525 minus 25 gives us zero. And since we're approaching five from the right, that's zero positive. And this means that we don't need to add a negative sign. Essentially five Z minus 25 is going to be a rewritten S five Z minus 25. Why? Well, essentially because the inner function is positive, we are approaching five from the right. And therefore, we can just drop the absolute value sign. So we can rewrite our limit limit when C approaches five from the right of five, C minus 25 divided by C squared minus 25. If we apply direct substitution, we get 25 minus 25 in the numerator which is zero and same in the denominator. So zero divided by zero is an indeterminate form because it is an indeterminate form. We want to eliminate that. And we can see that by noticing that in the denominator, we have a difference of squares. So we can factor out the difference of square says follows, we get limit when Z approaches five from the right in the numerator, let's factor out the five. And then in parentis, we end up with C minus five. For the denominator, we can factor it out as C minus five, multiplied by C plus five, right? Because we have a difference of squares. Our A is Z, our B is five. So we have Z squared minus five squared. Let's recall that A squared minus B squared as A minus B A plus B and in this case, A is ZB is five, right? And therefore we get C minus five multiplied by C plus five. Now, why is this useful while we can eliminate this hole in the rational function, which can potentially allow us to get the final limit as a finite number. So we are left with five divided by Z plus five. And now we can apply the rex substitution once again, which gives us five divided by five from the right plus five. And this is essentially five divided by 10, which can be simplified to one half. Since this is a finite number, we can state that we have successfully evaluated the limit. And this corresponds to the answer to ac one half. 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→2+ |2t − 4|t^2 − 4

Key Concepts

Limits

Absolute Value Function

Indeterminate Forms

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