Example of solving a limit that goes to infinity of a rational function. #calculus #limits
Пікірлер: 82
@harambe94616 ай бұрын
The biggest struggles in calculus is not even the calculus concepts it’s the algebra 😂 gotta get strong with the fundamentals before tackling something that takes it multiple steps forward.
@hebertengineering4 ай бұрын
Yes 100% correct.
@ihavenoenemis3 ай бұрын
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
@hebertengineering3 ай бұрын
@@ihavenoenemis Yes in the end limits are a lot about intuition.
@dowmanvarn7160 Жыл бұрын
The expression is NOT a rational function. A rational function is an expression that is the ratio of two polynomials. You have a radical in your expression, hence NOT a rational function.
@hebertengineering Жыл бұрын
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.
@dowmanvarn7160 Жыл бұрын
@@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.
@hebertengineering Жыл бұрын
@@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.
@chessandmathguy6 ай бұрын
@@dowmanvarn7160thanks, was looking for this! Yes a ratio of polynomials only. 🙂
@agooddoctorfan6515 ай бұрын
Okay then what would be the term for this?
@lovestein928 ай бұрын
im studying for finals tomorrow and u just saved me. THANK YOU
@hebertengineering4 ай бұрын
Happy to help!
@Inspirator_AG1128 ай бұрын
You can also omit the non-leading terms.
@Algebrainiac7 ай бұрын
You can just take the limit as x-> ∞ [sqrt(4x⁶)/-x³ ] and this gives us 2x³/-x³= -2
@snoyoclan25196 ай бұрын
I love you omg
@hebertengineering4 ай бұрын
You're thinking like an engineer!
@willisilikefood94922 ай бұрын
Or you can use Trocino Method and eliminate and only use the controlling factors.
@GauravkumarJaiswar8 ай бұрын
Thanks brother
@hebertengineering4 ай бұрын
No problem!
@sekhnicejulhas42193 ай бұрын
Respect❤
@khodieva_rano4 ай бұрын
Thx 😊
@hebertengineering4 ай бұрын
No problem 😊
@JM-ne5ep7 ай бұрын
YOUR VOICE MAKES ME WANT TO CLEARR MY THROAT AGAIN AND AGAIN WTF
@bangsimm6 ай бұрын
Fr
@hebertengineering4 ай бұрын
I'm going to take that as a compliment.
@IdrissChancy9 күн бұрын
😅😊🎉😂
@Delicious.Meiloorun6 ай бұрын
How is it 4 in top after multiplying x^3??? Would it be 4x^3???
@simonafrica-cn1ss5 ай бұрын
It's not multiplied by x ^3 , it's multiplied by 1 / x^3
@bruhh3759 Жыл бұрын
good
@ronscomic10 ай бұрын
Your voice... Your voice sounds like Rick in Rick and Morty
@hebertengineering9 ай бұрын
Only seen bits and pieces of Rick and Morty. I hope that's a compliment.
@uneducatedseinor49622 ай бұрын
And this should be the horizontal asymptote
@hebertengineering2 ай бұрын
Correct. Any limit approaching +/- infinity that evaluates to a finite number is a horizontal asymptote.
@RMF494 ай бұрын
I reason that the 1 and the 2 become irrelevant so get rid of them then simplify to -2
@hebertengineering4 ай бұрын
Nothing wrong with that logic! Now you're thinking like an engineer!
@geektoys37011 ай бұрын
Did it in my head in 10 sec
@Elcompalui71211 ай бұрын
How
@geektoys37011 ай бұрын
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
@elephantseal7816 Жыл бұрын
is it in indeterminant form when it is infinity/-infinity?
@mmmmmmok5292 Жыл бұрын
yes
@hebertengineering Жыл бұрын
Yes. Infinity/Infinity in any form (-/+ numerator or denominator mixed, etc.) is indeterminant form.
@nigel_w28818 ай бұрын
Saved me
@hebertengineering4 ай бұрын
Happy to help!
@seekingCK2 ай бұрын
did lhopital left us?
@tonatiuhdeleon82367 ай бұрын
Only applies to algebraic only polynomials
@rdxytshorts7778 ай бұрын
Where is the numerator x power 3
@hebertengineering4 ай бұрын
The square root of x^6 is x^3.
@douglaszare12153 ай бұрын
Your method fails if you take the limit of the same function to -infinity. You have to be more careful then that sqrt(x^6) = |x^3| not x^3.
@hebertengineering3 ай бұрын
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.
@douglaszare12153 ай бұрын
@@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.
@hebertengineering3 ай бұрын
@@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.
@douglaszare12153 ай бұрын
@@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.
@hebertengineering3 ай бұрын
@@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?
@user-rx6bp8zf2j5 ай бұрын
1
@neko628012 ай бұрын
Oh wait this is so easy
@zayed990210 ай бұрын
why you not simplify the root It will be 1+2x^3 You have 2x cube up and -x cube down Use the rule and Divide to get -2 Don’t miss the absolute
@vanionings9 ай бұрын
Einstein
@sumantrathia89456 ай бұрын
+ool9
@hebertengineering4 ай бұрын
Thinking like an engineer!
@Uskebasa Жыл бұрын
This is not correct at all. The limit is -2 only when the x approaches +infinity. Otherwise, when x approaches -infinity the limit goes to +2. You can see it by graphing the function or you can factor 4x^6 inside the square root and then bringing the factor out leaving with |2x^3|. So, since we have to different values the limit as x approaches infinity does not exist.
@hebertengineering Жыл бұрын
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).
@Uskebasa Жыл бұрын
@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.
@hebertengineering Жыл бұрын
@@Uskebasa Interesting
@targetstudies9007 Жыл бұрын
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 !
@WalkinWhiskey11 ай бұрын
@@hebertengineeringits okay you’re right no sign implies its positive
@bigbungus486911 ай бұрын
L'hopitals rule
@user-yn9xd3my1q10 ай бұрын
No
@vanionings9 ай бұрын
nah
@alanhernandez35414 ай бұрын
You went tooooo fast
@hebertengineering4 ай бұрын
These videos are supposed to be fast. Check out our long form content. We have full length videos with many limit examples done slowly.
@JM-ne5ep7 ай бұрын
STOP SMOKING
@hebertengineering4 ай бұрын
I don't smoke so...
@duf29 ай бұрын
why is the square root of 1 +x^6 is the same as x^3????
@hebertengineering9 ай бұрын
The square root of 1 over x^6 is the same thing as 1 over x^3.
@duf29 ай бұрын
@@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
@hebertengineering9 ай бұрын
@@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.
@duf29 ай бұрын
@@hebertengineering okay thank you! I hope your day is going well like you made mine 10x better