evaluate-the-function-at-01001-and-0001-from-both-sides-of-the-value-it-approaches-in-exercises-53-8

Question

Evaluate the function at $0.1,0.01,$ and $0.001$ from both sides of the value it approaches. In Exercises $53$ and $54,$ evaluate the function for values of $x$ of $10,100,$ and $1000$ From these values, determine the limit. Then, by using an appropriate change of form, evaluate the limit algebraically and compare values.
$$\lim _{x \rightarrow \infty} \frac{x^{2}+5}{\sqrt{64 x^{4}+1}}$$

Anish Wadhwa · Accepted Answer

suppose we have this function X times natural log of X and we want to find the limit as excess approaching zero from the right of this function. So we&#x27;re only going to be dealing with values to the right of zero. In other words, we&#x27;re only going to be dealing with Excel use that are greater than zero. So we will be using values that really, really close to zero for this. So if you were to calculate these values, you would find that when x is equal 2.1 f of X equal to negative 0.230 to six when X equals 0.1 F of x is equal to negative zero point zero for 605 When I says equal to 0.1 Why is equal to negative? Is hero point 00 69 one? You can already kind of see a trend as we&#x27;re continuing this so we can kind of hypothesize a little bit of water limit is going to be, but let&#x27;s keep on going. Our next values are negative zero point 0009 to 1 in our final values, negative 0.115 And we&#x27;re actually going to use just that value, and we&#x27;re going to be rounding this up to two now. If we look at where zero is, which will be down here. If we see this trend, we see that it&#x27;s that our values air increasing from a neighbor from a really small negative to an even smaller negative. So they&#x27;re increasing but still negative, right? And if we see that our values are getting more zeros in them, if we get a number, say, like point zeros 0000001 or like so on and so forth, we get approximately something like 0.1 or something like that, right? It will be a very a very small number, and we can use this to kind of approximate that. When that are limit as X approaches, zero from the right will be well, zero

Foster Wisusik · Answer

Okay, This question wants us to find this limit. So in order to use our limit rules, we&#x27;re going to have to first, right? This is a fraction. So let&#x27;s begin by multiplying by the conjugated of this expression on top and bottom. So we take our original expression right here. And let&#x27;s just put the experts one group that together because now we can see the conjure. Git would just be that expression. But this time it&#x27;s split with a minus site. So now we can just begin to simplify. So expanding out the numerator Now we get just the difference of the squares of these two terms. So the square root goes away and then we subtract the square of X plus one and our bottom stays the same. And now we can compute this limit. So the first thing we should do is expand out the top. So expanding out that&#x27;s out. While expanding out the square, this becomes X squared plus two x plus one. So subtracting we get zero in the X squared term and then we get a five x minus two x so three X and then for the constant, we get five minus one, which is four. And then we&#x27;re still dividing by our bottom expression. So now what we need to dio is look at the terms that will dominate. So on top, this now will behave like three X and then on the bottom it&#x27;ll behave like square root, X squared minus X. And then if we take that, that square X squared becomes absolute value X And since we&#x27;re dealing with negative numbers in this limit absolute value, X just behaves like negative X. So we get three x over negative two x or negative three halves for the limit.

Bahar Tehranipoor · Answer

So in this exercise, we want to evaluate this limit. And to do this, we&#x27;re gonna actually use a trick. So we&#x27;re going to say this limit equals L into a value intel. We&#x27;re gonna take the Ellen of both sides. So we take the Ellen of both sides. We have the limit of ex approaching infinity ln of this x over X was one of the X power. So before I write anything else, only have to remember that when we work with logarithms and l s, there&#x27;s actually a rule that if we have a power, so just here we have to the x power we could actually bring that down and next to the l s, we could be right this as ex ln of X over X plus one. This is gonna equal the ln a belt. So now, to find out, we can first evaluate the l N f o and then come back, take heed of both sides of what we find, and then we&#x27;ll have the limit that we&#x27;re looking for. But right now we could just evaluate for the ln of the limit we&#x27;re looking for, because with this trick, we could bring the power down and it&#x27;s a lot easier to evaluate, so we can actually use will be taught here. We could rewrite it as a limit of ex approaching infinity of the ln of X over X plus one. And we can have that over one over X because multiply my ex is the same as dividing by one over X. Now, to make sure we can use will be tall. We have to see if our conditions are true. So we have to have the numerator and denominator or both defensible, which they are. And we also have to have that the, uh if we have experts affinity for both the numerator denominator, we&#x27;re gonna have something indeterminate. So that is true. Because if we have ex approach infinity of what&#x27;s inside this Ellen, we&#x27;re going to get one. So we have Ellen of one, which is zero. And then in our denominator, we have won over something approaching infinity, which is also zero. So zero over zero is indeterminate, meaning we can use little bit top. So we&#x27;re gonna call this top function f of X, call this bottom function g of X and take the derivatives will do right here. Derivative of F of X and derivative of Kiev X. So this is actually pretty tricky derivative of June exes. Pretty easy, because that&#x27;s just gonna be negative. One over X squared, but derivative f of X. Now, that is gonna be a little So enter the derivative of anything ln is gonna be won over that thing so we&#x27;ll have one over this X over X plus one. Now, what we have to do is multiplied by the derivative of what&#x27;s inside this ln this derivative of X over X plus one. So let&#x27;s go to the next patient. Actually, just work on that. So we want to find the derivative of X over X plus ones. This is gonna be pretty tricky now. What we can do is we can actually just use either the problem of the potion rule. I&#x27;m going up for the product world because I think that&#x27;s easier, because right now we have X Times X plus one to the negative one power. We could do the quotient rule with experts with me the denominator, of course, but I just think it&#x27;s easier to use this. So using the product roll. We&#x27;re gonna keep the first term ex multiplied by the derivative the second term. So most of my mind of this we have. We bring the power down negative one times X plus one Subtract one from the power. So we have to the negative to power. Now we have plus keep the second term X plus one to the negative one and take the dirt of the first term, which is just one. So that&#x27;s all we gotta do now. Let&#x27;s just clean this up. Will have negative X over X plus one squared plus one over X plus one. Now we can actually multiply both the numerator and denominator here by experts one so that we can have a common denominator. So we&#x27;ll have that&#x27;s plus one here. Now we can add them. So if we add that we get the denominators X plus one squared and then the numerator have negative X plus X plus one. So we just have one. So that means we&#x27;re multiplying here by this derivative right here. Which iss this one over X plus one square. Let&#x27;s do that. We have one over X plus one switches is pretty messy now. Let&#x27;s vote to 1/3 page so it will copy all this down. So we&#x27;ll have now will be the limit of expert in infinity Moment of ex approaching infinity with our derivative of F on top. So we will have one over X over experts. One is the same thing as experts. One over X. We could just write that Well, that&#x27;s plus one over X Times artery that we took here, which is just one over X plus one squared. You can actually come back. It&#x27;s okay by this in a second and then we have over negative one over. Expert we could have won over. What we could do is we can actually cross out one of these experts ones because we have experts one and then over. Experts want square so we can remove this. Remove this power right here and we write this one more time as just one over x Times X plus one. I think this is over negative one over X squared. Now let&#x27;s try and clean up again. If you have over a negative one over experiment actually moved by that x squared and so we can have. Let&#x27;s see, we have won over at times X plus one in normal, deploying by negative X squared. You could cancel one of the excessive denominator Take this power away. So we&#x27;re really just looking for the limit of ex approaching infinity of negative X over X plus one just like that. Now, if we have expert infinity, we have that we&#x27;re going to have negative one. That&#x27;s it. It&#x27;s just gonna equal negative one, because this one constant right here, we can pretty much ignore it because, as ex gets really, really big, this won&#x27;t really make a difference. We could think of it as just negative x over X and negative X over exit. Just maybe one, you know might be really easy to just go ahead and say, Oh, our answer&#x27;s negative one. That&#x27;s not You have to go back to the beginning and see that we actually found ln out And we were really looking for l. So we know that Ellen of El Equal lt&#x27;s negative one. And to find the real limit, we gotta do either both sides. So we have to be to the negative one, which we could also say Our answer is one over so that&#x27;s a pretty long problem. But using this trick by taking the eligible sides of our limit, that is how you solve it.

Bahar Tehranipoor · Answer

in this problem. We want to evaluate this limit. And this was an interesting one, because we&#x27;re gonna have to use a trick. So we&#x27;re going to say that this limit is equal to l like that. Any very well call X just call it. Actually, Caves X was already using that. But like l m whatever, that&#x27;s gonna be the value of what our limit is going to evaluate too. So to make it easier for single limit, we&#x27;re actually going to take the Ellen of both sides. So we have the limit of ex approaching infinity of the ln of extra one over X Square is going to equal the ln off what we&#x27;re looking for are elsewhere looking for. So what we can do now bring that one over X square down. That&#x27;s just a property of loss and logarithms so we can bring it down. So we have Elena. That&#x27;s over F Square. That&#x27;s what we&#x27;re gonna live it up. So if we find this limit, you can take you to that power and then we&#x27;ll have our original limit. So let&#x27;s do that. Speaking use will be tell here calling this stuff of ex calling this G of X is gonna be a determinate. If we have expert community, so we can use will be tough. We&#x27;ll take the derivative. A lot of Exeter of it is one of our X derivative of X squared is just going to be two x. So we&#x27;re looking for the limit of ex approaching infinity off one over to that square here, and that&#x27;s going to equal zero. So don&#x27;t stop there. Don&#x27;t forget that work you just found ln of what we&#x27;re looking for. We just found ln off our original limit. So we have to do now if we have to raise, be to what we just found, even zero. And that means l which is what we&#x27;re looking for. Our answers actually won because l a seat is a zero. We found Eleanor, which is zero. Either zero gives us one, which is our original limits value. So this is a really cool trick you can use where we have something to a power that has that a variable accident. We can use a little of both sides of our limits and people to bring it down. So it&#x27;s easy to violet, So why do it

Jason Orozco · Answer

I want to find the limit as X goes to zero of the function X divided by Route 16 X plus one minus one all to the power of 1/3. Um, so first will just use the limit rule for roots up. Teoh Rewrite. This adds parentheses limit as X goes to Ciro X Group 16 X plus one. But it&#x27;s one, and we&#x27;re raising the entire limit to the 1/3 power. So that allows us to sort of disregard that, uh, 1/3 exponents until after we brought you the inner limit. Uh, so to do that, since we have Well, first we want toe check. Direct substitution. We can see that if we plug in zero into the denominator, we&#x27;re gonna get route 16 times their A plus one minus one. Uh, which is route one minus one, which is zero. So direct substitution is not going to work. Um, you to try something else? So since we have a radical, uh, minus one, we&#x27;re going Teoh, let&#x27;s multiply at the top of the bottom by the contra kit. So that&#x27;s going to give us get the parentheses going, uh, limit as X goes to zero of X times and then the contra git is route 16 x plus one plus one changed sign divided by, um, and then multiplying that that, uh, the same expression by the original denominator, the radicals gonna cancel out. So we&#x27;ll get 16 x plus one, and then the minus one times plus one will give us minus one so the ones will cancel out. So the denominator is just 16 X. Uh, so this expression is still equal to our original limit. And now this is a zero, Um, and so we can see that we tried direct substitution here plugging zero and gives us zero in the denominator. So that&#x27;s not gonna work. Um, but we do have an X, and he has a factor of the numerator and the denominator. So if we just look at the left or right limit by itself. So let&#x27;s say we just look at the limit as X goes to zero plus well, on the right, X is not equal to zero. Which means that if we wanted to find the right limit of this expression, we could weaken, cancel out the exes from the numerator and the denominator and get six, Route 16 x plus one plus one divided by 16. Um, and so this is this no longer has a serum denominator because he has 16 so we&#x27;ll just, uh, used directs. Institution plug zero into X 16 times zero plus one plus one worker 16. Um, and that gives us to over 16 war 1/8. Um, so we can see that if we were to take clear to change this to the limit as excuse to zero minus, we could apply the exact same calculations. Excellent. Cancel out, because X isn&#x27;t zero to the left of zero. Um, and then we would use direct substitution. So the limit as X goes to zero minus is also 18 since the left and right limits are equal. Um, we can conclude that the original limit as X goes to zero up here is also equal to one. It just by definition, um, so, uh, continuing from up here, we can plug in 1/8 for the limit, and it&#x27;s still raised to a 1/3 power, and that gives us 1/2. So our limit is excuse to zero is just 1/2

Problem

Solve the given problems involving limits. Evalua…

Problem 54 Easy Difficulty

Answer

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 1 / AB Educators

Anna Marie V.

Caleb E.

Kristen K.

Samuel H.

Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 23

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Anna Marie V.

Caleb E.

Kristen K.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Basic Technical Mathematics with Calculus

Anna Marie V.

Caleb E.

Kristen K.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Anna Marie V.

Caleb E.

Kristen K.

Samuel H.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus