Yes 100% correct.

I think just from assumption, here it work like this when x approach infinity there are lot of term in the equation which we can neglect, so dividing by the highest power gives the total ratio of how the function will grow

@@ihavenoenemis Yes in the end limits are a lot about intuition.

Dowman Varn thank you for your input and for the clarification. The evaluated limit is correct, however, and the intuition behind evaluating these limits I think is illustrated.

@@hebertengineering The evaluated limit is correct and the method is fine. I hope no one would take from comment a suggestion otherwise. It is just that there was an implication that the function was a rational function, and it isn't. I think that it is important to get the terminology correct.

@@dowmanvarn7160 Agreed. I'm going to do the best I can to make all of the content on this channel as polished as possible. Every problem posted, however, will have the correct answer determined with a correct method.

​@@dowmanvarn7160thanks, was looking for this! Yes a ratio of polynomials only. 🙂

Okay then what would be the term for this?

Happy to help!

I love you omg

You're thinking like an engineer!

No problem!

No problem 😊

Fr

I'm going to take that as a compliment.

It's not multiplied by x ^3 , it's multiplied by 1 / x^3

Only seen bits and pieces of Rick and Morty. I hope that's a compliment.

Correct. Any limit approaching +/- infinity that evaluates to a finite number is a horizontal asymptote.

Nothing wrong with that logic! Now you're thinking like an engineer!

How

see, i simply " ignore" the Constants because at infinity they are much smaller and unimportant near the x squared.. then i take the root and cancel

yes

Yes. Infinity/Infinity in any form (-/+ numerator or denominator mixed, etc.) is indeterminant form.

Happy to help!

The square root of x^6 is x^3.

As x goes to - infinity the you still end up with two x terms that approach 0. The limit approaches -2 as x approaches both positive and negative infinity.

@@hebertengineering Are you seriously saying the limit of this function is -2 as x goes to -infinity? That is wrong, as others have pointed out, too, and it shows that your method is incomplete or misleading. Rational functions have the same limit at +infinity and -infinity but that is not necessarily true for functions with radicals like this. Try plugging in some values with x very negative. The numerator is always positive. The denominator is positive when x is negative (in fact, less than 2^(1/3)). So, you claim the ratio between positive numbers approaches -2? For example, when x=-10, the function is sqrt(4,000,001)/1002 = 1.996. When x=-100, the function is 1.999996.

@@douglaszare1215 ​​⁠ Just took a closer look. I see so for example the original function as x goes to - infinity the denominator becomes positive. x^6 always results in a positive number. So the limit as x goes to -infinity is 2. Don’t think properly evaluating principalroot(x^6) as abs(x^3) would fix the problem either bc the method would still say the limit is approaching -1 in the denominator. I’ll take a closer look at the comments when I can and consider posting a short and/or long form video about this.

@@hebertengineering x/|x| can be called sign(x). Then sqrt(4x^6+1) = x^3*sign(x)*sqrt(4+1/x^6) and you can cancel the x^3 leaving sign(x)sqrt(4+1/x^6) in the numerator and -1+2/x^3 in the denominator.

​@@douglaszare1215 Ok I see. You factor out an x^6 from the expression inside the root. The sqrt(x^6) evaluates to abs(x^3). Then abs(x^3)/(x^3) is just sign(x). I've worked a lot of calculus problems and unfortunately never ran into this issue yet. May need to revise my limits videos accordingly. Does the method shown here work for general rational functions, i.e., polynomial divided by polynomial?

Einstein

+ool9

Thinking like an engineer!

No minus sign in front of the infinity symbol implies positive infinity. So the question is the limit as x approaches positive infinity of the given f(x).

​@hebertengineering Sorry, but I don't agree. Impling a sign could only lead to misinterpretation. In my country, for example, we always tend to explicit the sign.

@@Uskebasa Interesting

No that's wrong , -infinity actually represents the infinity -h , where h tendings to zero ,which is nothing but a positive number , answer is accurate . In india during our jee preperation we literally do warm up with these questions !

@@hebertengineeringits okay you’re right no sign implies its positive

No

nah

These videos are supposed to be fast. Check out our long form content. We have full length videos with many limit examples done slowly.

I don't smoke so...

The square root of 1 over x^6 is the same thing as 1 over x^3.

@@hebertengineering but why? But its 1/x^6 if I take it under the root it would be -x^3? Can you explain it in details please and thank you for your video

@@duf2 Write down 1/x^6. Then put the entire 1/x^6 expression under a square root symbol. Next, you can separate the square root into both the numerator and denominator. In the numerator, the square root of 1 is just 1. In the denominator, the square root of x^6 is x^3. Not sure why you are thinking you get -x^3. Perhaps you are thinking that 1/x^6 can be written as x^-6. The square root of x^-6 can be written as x^(-6/2) which is x^-3. x^-3 can be written as 1/x^3.

@@hebertengineering okay thank you! I hope your day is going well like you made mine 10x better

@@duf2 No problem! Happy to help.