💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!



Numerade Educator



Problem 23 Easy Difficulty

Find the limit or show that it does not exist.

$ \displaystyle \lim_{x \to \infty}\frac{\sqrt{1 + 4x^6}}{2 - x^3} $



More Answers


You must be signed in to discuss.

Video Transcript

All right. We want to evaluate this limit or show that does not exist. Um Some people will talk about strategies and limits like low petals rule or evaluating and behavior. Local tales rule isn't great because limits have to exist before derivatives evaluating and behavior can be hand wavy or confusing. Um So I like to do it with algebra. Um so the problem with this is that the top is really big, right? We have really big to the power of six and then some confusing stuff going on and the bottom is also really big. Probably negative, really big. Um and we don't know which one is bigger if any of them are. So I'm going to stop the things from being big and what I'm gonna do is divide top and bottom by X cubed Y X cubed because inside the root the root of X. The sixth is X cubed. So um so once I bring the, so here I'll do this, I'm going to write one over X cubed in the top, I'll deal with that in a second. And then in the bottom I'm going to just divide by execute. So this is two over X cubed minus execute over execute as one. Okay so what happens is this x cubed can go inside the root but I'll be dividing by extra six now and then uh infinity is positive. So X is a really big positive numbers so I don't have to worry about weird behavior inside roots. Sometimes when X is negative you do and that's that can be kind of weird. So oh sorry I'm dividing by extra six. I said make it a little more clear that I'm not writing that Near that. Okay, X to the 6th Plus four. Extra 6 over extra six. Just 4. Okay Now this is two x. I'm gonna write actually I'm gonna write this is one over X to the 6th. I'm going to write this is two times one over X. To the six minus one. And now the only theory my need for limits when something is getting big is that one over something gets small, one divided by a huge number is zero. So now I'm going to use that here, I'm going to put everywhere. I have one over X. Sorry you let me make a mistake I guess this is a video. So you're not allowed to yell at me. Usually when I'm teaching live you can yell at me. Okay, I'm gonna put in zero for one over X everywhere. Yeah. So then the top is just going to be the root of really small thing to the six is still really small. Plus four. So rude of four is two and then bottom two times really small thing cube, that's still all zero -1, so this is -1, so he get minus two. And I think that's the cleanest way to think about a problem like this.