Limit at infinity

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Күн бұрын

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@junchen9954 Жыл бұрын

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 Жыл бұрын

Totally agree.

@epikherolol8189 8 ай бұрын

But this is highschool level in my country

@hafizusamabhutta 7 ай бұрын

I don't like when he does the easiest steps

@pimp88ziengs Жыл бұрын

I studied math at uni and this video is amazing 👍. The calm, clear and concise explanation is nice.

@hipepleful Жыл 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 Жыл бұрын

inf * inf = indeterminate form

@hipepleful Жыл бұрын

@@waltz251 no its not. It's just infinity.

@epikherolol8189 8 ай бұрын

@@hipeplefulYeah

@twinkletoes1588 Жыл 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

@beezy8551 9 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 9 ай бұрын

Goodluck

@winonavalentina6206 3 ай бұрын

Instantly subscribed, i love how you explain things so slowly cuz im a fricking idiot, THANKYOU

@SuperTommox 11 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)

@zeynepartut3362 2 ай бұрын

By doing the thing at 7:38 you just made my day.

@Samuel-cl1cv Жыл 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.

@cherryisripe3165 Жыл бұрын

Very nice problem and brilliant explications. Thanks a lot

@saharashara7980 Жыл бұрын

شكرا استاذ تعلمت منك الكثير

@malikahashami Жыл 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.

@alanwebber238 11 ай бұрын

To infinity, and beyond . . .

@alanwebber238 11 ай бұрын

#BackToTheFuture

@snowman2395 7 ай 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

@mom-MA1 2 ай бұрын

wow, i appreciate that [the tricky -] , hero.

@epikherolol8189 8 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 +∞

@KingGisInDaHouse Жыл 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

@مسعودکشاورز-ب2ص 10 ай бұрын

Very nice thank you.

@vishalmishra3046 3 ай бұрын

Here limit of positive vs. negative infinity give different results. *That's interesting and unexpected* Positive case Take 2 x^7 common to get 2 x^7 [ ( 1 + x^-7 / 4) ^ (1/2) - 1 ] which binomially simplifies to 2 x^7 (1 + 1/8 x^-7 - 1) since 1/x tends to zero = 2/8 = 1/4 Negative case x is negative, so x^7 is negative but 4 x^14 is larger positive than x^7 whose positive root is 2 x^7, so the final result tends to a sum of 2 positive numbers, 2 N^7 + 2 N^7 = 4 N^7 = *Positive Infinity*

@vishalmishra3046 3 ай бұрын

To eliminate any surprise from negative infinity in limit problems, it is best to transform the problem into *positive infinity* problem by changing variables to using y = -x and then solve the expression in Y space.

@YingchenHe-k8s 8 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

@user-pl7tr9dv6l 10 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 ...

@kellesonekellesonekelleson2405 Жыл бұрын

What background is this?

@emilie375 Ай бұрын

I didn't get why this that not worked: - lim x->- infinity (4x^14 + x^7) = lim x->-infinity 4x^14 = +infinity (by factoring by x^14) - lim t-> +infinity sqrt(t) = + infinity so by composition lim x->- infinity sqrt(4x^14 + x^7) = +infinity - lim x->- infinity (-2 x^7) = + infinity Thus, by sum the limit is +infinity. I mean, ok with infinty-infinity we can't conclude but intinity + infinity is infinity, isn't it ?

@sunil.shegaonkar1 6 ай бұрын

At 9 minutes square root of a positive number is positive, so why change sign? Because limit tends to minus infinity should not affect the sign of under root factor. We did not apply any limiting values until 9 minutes, so √x power 7 should be treated as positive in my opinion.

@Bruce-oc9hf 9 ай бұрын

well i have diveded by x to the power of 7 and result was -4 . any explination

@user-pl7tr9dv6l 10 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 10 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 10 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 10 ай бұрын

Yes. I realized it after I responded. And the graph actually shows ¼ for positive infinity.

@gp-ht7ug 10 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

@skwbusaidi 8 ай бұрын

I always prefer to change limit to -inf to inf , specially when there is square root in the problem

@anatolysolunin4991 11 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 11 ай бұрын

I’m sorry, I was wrong. The value in root less than 4, because of odd degree of one of terms.

@onepiece3mk323 6 ай бұрын

I mean , just plug the infinite and its done . Am i wrong ????

@hemangkulkarni3947 Жыл бұрын

I believe this answer is wrong i checked with both rationalising and a different method

@PrimeNewtons Жыл бұрын

@abderrahmanwakrimi Жыл бұрын

La réponse est évidente. +infini-(-infini)=+infini.

@حامدعبدالفتاح-ت1ض Жыл бұрын

Oh man, the limit equals =∞

@ynag1223 Жыл бұрын

this is difficult to learn ±∞ and ±√ combo

@dmihovilovic Жыл 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.

@tomasriquelmeleroy1716 Жыл бұрын

Who else is 17???

@李家生-j2v Жыл бұрын

1/4

@FrancisHealy-w9f Жыл бұрын

This is wrong. The result converges to 0.25. It is incredibly easy to prove this.

@bhaskarporey3768 Жыл бұрын

Then prove it.

@sebas31415 Жыл бұрын

Prove it

@khrysztoffe27 Жыл 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.