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.
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
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.
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.
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?
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.
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.
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 !
@@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.
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.
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.
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.
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?
im studying for finals tomorrow and u just saved me. THANK YOU
Happy to help!
You can just take the limit as x-> ∞ [sqrt(4x⁶)/-x³ ] and this gives us
2x³/-x³= -2
I love you omg
You're thinking like an engineer!
You can also omit the non-leading terms.
ok
Or you can use Trocino Method and eliminate and only use the controlling factors.
Respect❤
good
Thanks brother
No problem!
Can’t we just use the shortcut since they technically have the same degree and do 2x^3/-x^3 and get our answer that way
How is it 4 in top after multiplying x^3??? Would it be 4x^3???
It's not multiplied by x ^3 , it's multiplied by 1 / x^3
is it in indeterminant form when it is infinity/-infinity?
yes
Yes. Infinity/Infinity in any form (-/+ numerator or denominator mixed, etc.) is indeterminant form.
Thx 😊
No problem 😊
And this should be the horizontal asymptote
Correct. Any limit approaching +/- infinity that evaluates to a finite number is a horizontal asymptote.
Your voice... Your voice sounds like Rick in Rick and Morty
Only seen bits and pieces of Rick and Morty. I hope that's a compliment.
Did it in my head in 10 sec
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
I reason that the 1 and the 2 become irrelevant so get rid of them then simplify to -2
Nothing wrong with that logic! Now you're thinking like an engineer!
did lhopital left us?
Calc is easy I just suck at the algebra part when it comes to complex problems
Saved me
Happy to help!
Where is the numerator x power 3
The square root of x^6 is x^3.
😅😊🎉😂
Only applies to algebraic only polynomials
YOUR VOICE MAKES ME WANT TO CLEARR MY THROAT AGAIN AND AGAIN WTF
Fr
I'm going to take that as a compliment.
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
Einstein
+ool9
Thinking like an engineer!
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.
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?
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.
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
1
Oh wait this is so easy
L'hopitals rule
No
nah
You went tooooo fast
These videos are supposed to be fast. Check out our long form content. We have full length videos with many limit examples done slowly.
STOP SMOKING
I don't smoke so...
why is the square root of 1 +x^6 is the same as x^3????
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.