I really love how you're doing college level math while approaching everything like you're speaking in front of a bunch of 6th graders. Not the kind of patience and calmness everyone has.
@bobwineland9936
8 ай бұрын
Totally agree.
@epikherolol8189
2 ай бұрын
But this is highschool level in my country
@hafizusamabhutta
18 күн бұрын
I don't like when he does the easiest steps
@pimp88ziengs
8 ай бұрын
I studied math at uni and this video is amazing 👍. The calm, clear and concise explanation is nice.
@hipepleful
8 ай бұрын
I did it by factoring. Ignore the square root, and just focus on whats inside: 4x^14+x^7. Factoring makes it become x^7(4x^7+1). This goes to -inf * -inf, which is just positive infinity. Square rooting a positive infinity still equals infinity. Now, to deal right the -2x^7. It is -2*-inf. This is also + infinity. Infinity + infinity = infinity. This, the entire limit is infinity.
@waltz251
6 ай бұрын
inf * inf = indeterminate form
@hipepleful
6 ай бұрын
@@waltz251 no its not. It's just infinity.
@epikherolol8189
2 ай бұрын
@@hipeplefulYeah
@Samuel-cl1cv
8 ай бұрын
I don't know if I'm right but here is how I did it: You can guarantee that the expression in the square root will be greater than 0 because x^14 grows faster than x^7. So, the result of the square root will be a positive number. As x goes to negative infinity, -2x^7 goes to infinity. Therefore, when x goes to infinity, the whole expression goes to infinity.
@twinkletoes1588
8 ай бұрын
You are awesome sir! I'm now very intersted in math after your videos! I can't speak your language, but try my best to understand everything you explain
@cherryisripe3165
5 ай бұрын
Very nice problem and brilliant explications. Thanks a lot
@beezy8551
3 ай бұрын
have my Calculus exam tomorrow and your videos are so consice and easy to understand. And the way youn explain somehow calms my nerves😂.Your videos have come in clutch
@PrimeNewtons
3 ай бұрын
Goodluck
@saharashara7980
8 ай бұрын
شكرا استاذ تعلمت منك الكثير
@SuperTommox
5 ай бұрын
I know how to solve these limits but you always find some ways i don't know. Love it! Especially the way to solve the irrational function towards the end. I would have used a change of variable: y=(-x) So that we have: Lim(y->+inf)
@user-rq3tl1nk8r
3 ай бұрын
Very nice thank you.
@malikahashami
8 ай бұрын
In the square root it's probably possible to factorise by x^7. This way we can get it to become the sqrt of x^7(4x^7+1) which is plus infinity ,and by adding -2x^7 the result will still be the same mayby.
@snowman2395
25 күн бұрын
despite being a pre-calc student im really good at guessing these limits and its mostly looking at which x is 'bigger' even tho there both inf
@kellesonekellesonekelleson2405
8 ай бұрын
What background is this?
@alanwebber238
5 ай бұрын
To infinity, and beyond . . .
@alanwebber238
5 ай бұрын
#BackToTheFuture
@KingGisInDaHouse
8 ай бұрын
You can argue that the 14th power is a higher order infinity. Or that the end behavior of the polynomial would tail off at infinity on the right side. Delegating that to the reader, Lim x->-inf sqrt(4x^14+x^7)-2x^7 Adding a constant in the radical isn’t going to make a difference As the sqrt contains a higher number adding some change doesn’t make a difference sqrt(4x^14+x^7+1/16)≈sqrt(4x^14+x^7) lim x->-inf sqrt([2x^7+1/4]^2)-2x^7 |2x^7+1/4|-2x^7 |-inf|-(-inf)=inf
@user-pl7tr9dv6l
3 ай бұрын
To all of you who suggested 0.25 as the limit (like I did): that's indeed the limit when x goes to infinity - but not the limit when x goes to negative infinity, as the problem had it 😉 like my math teacher said: "take time to read the problem text carefully". Which I clearly didn't ...
@epikherolol8189
2 ай бұрын
We can solve this without any calculations too. x¹⁴ is +∞ while x⁷ is -∞ BUT x¹⁴ term's magnitude will be way larger than x⁷ term's magnitude, so we can neglect x⁷ term from square root. After neglecting, only square root of 4x¹⁷ - 2x⁷ remains. Square root of 4x¹⁷ is -2x¹⁷ So the result will be -4x¹⁷ After putting x as -∞, we get ∞+∞ situation, so thst will turn out to be +∞ only. So the answer is +∞
@user-op2sb4vm2f
Ай бұрын
Hi, I have an idea.When x approche -Infinity, √(4x^14+x^7) is approximately =√4x^14.Since √4x^14 is positive, so √4x^14=2 (absolute value x )^7, which is positive infinity, and if we -2x^7, it equals to4(absolute value x)^7, which is also positive infinity. I think this way is more simple
@Bruce-oc9hf
2 ай бұрын
well i have diveded by x to the power of 7 and result was -4 . any explination
@skwbusaidi
2 ай бұрын
I always prefer to change limit to -inf to inf , specially when there is square root in the problem
@gp-ht7ug
3 ай бұрын
I would have solved it like this: let X>0 then I would have divided the radicand by x^14. Then the limit would have been |/(4+1/x^2) - 27x^7 => sqrt(4+0) - 27^-inf > 2+inf => +inf
@user-bf4yw9qx5i
8 ай бұрын
Oh man, the limit equals =∞
@user-pl7tr9dv6l
3 ай бұрын
Hello Prime Newton, I like your channel! But, you got this limit wrong. One can show that the entire expression never exceeds 1/4 by first just writing the expression under the square root as (2x^7 + 1/4)^2 - 1/16, and then bounding the entire expression from above, as follows: √(4x^14 + x^7) - 2x^7 = √((2x^7 + 1/4)^2 - 1/16) - 2x^7 < √((2x^7 + 1/4)^2) - 2x^7 = 2x^7 + 1/4 - 2x^7 = 1/4 You can actually "see" the limit as well from the modified expression: for large X, the square expression (under the square root) dominates and the impact of subtracting 1/16 is completely negligible, that is, for large X: (2x^7 + 1/4)^2 - 1/16 ≈ (2x^7 + 1/4)^2. This already 'hints' the limit is 1/4. This can also be shown by straighforward algebraic manipulation and use of l'Hospital's rule: first pull out 2x^7 from under the square root, factor the resulting expression (with 2x^7 is a common factor), and then 'flip down' the 2x^7 as 1/2x^7 into the denominator. The resulting expressions tends to "0/0" when x->∞ which means you can apply l'Hospital's rule, and get 1/4 as the limit.
@PrimeNewtons
3 ай бұрын
I remember a similar comment I read about this limit problem. I am pretty sure my steps are correct and my answer is also correct. The graph of the function also confirms the answer. Infinity could be a little tricky. I have had to take down some of my videos because I played with infinity 😭.
@user-pl7tr9dv6l
3 ай бұрын
Ah, you're right. I accidentally took the limit as x goes to infinity, and not to 'negative infinity'. I'm pretty sure the others who suggested 0.25 as the limit did the same thing.
@PrimeNewtons
3 ай бұрын
Yes. I realized it after I responded. And the graph actually shows ¼ for positive infinity.
@user-st3ej6bi9b
7 ай бұрын
1/4
@onepiece3mk323
3 күн бұрын
I mean , just plug the infinite and its done . Am i wrong ????
@user-nk4vd3ow5x
7 ай бұрын
this is difficult to learn ±∞ and ±√ combo
@anatolysolunin4991
5 ай бұрын
I’m afraid, the result is not plus infinity. The value under the square root is little more, than 4. So, the square root of it is a little more, than 2. And the bottom tends to 0 from the left, but not from right!
@anatolysolunin4991
5 ай бұрын
I’m sorry, I was wrong. The value in root less than 4, because of odd degree of one of terms.
@abderrahmanwakrimi
5 ай бұрын
La réponse est évidente. +infini-(-infini)=+infini.
@hemangkulkarni3947
7 ай бұрын
I believe this answer is wrong i checked with both rationalising and a different method
@PrimeNewtons
6 ай бұрын
Ok
@tomasriquelmeleroy1716
8 ай бұрын
Who else is 17???
@dmihovilovic
6 ай бұрын
Too complicated explanation. Instead sqrt(4x^14+x7) -> 2|x|^7 when x -> -infinity. The second part -2x^7 will go towards +2|x|^7 when x -> -infinity because of the sign. Therefore both together will go to +infinity.
@user-ud1zv2yh3r
8 ай бұрын
This is wrong. The result converges to 0.25. It is incredibly easy to prove this.
@bhaskarporey3768
8 ай бұрын
Then prove it.
@sebas31415
7 ай бұрын
Prove it
@khrysztoffe27
6 ай бұрын
He's correct and you're wrong. Use your calculator to find out that the answer of the limit is +inf regardless of the value of x in the -inf. He proved it perfectly that 1 divided by any number approaching to 0 from the right is +inf.
@josefranciscogarduno5278
6 ай бұрын
You are wrong. Correct answer is 1/4
@PrimeNewtons
6 ай бұрын
Ok
@khrysztoffe27
6 ай бұрын
He's correct and you're wrong. Use your calculator to find out that the answer of the limit is +inf
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