Limit at infinity

In this video, I showed how to evaluate a survd limit at negative infinity

[junchen]

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.

[pimpziengs]

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

[hipepleful]

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.

[zeynepartut]

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

[twinkletoes]

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

[Samuel-clcv]

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.

[winonavalentina]

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

[SuperTommox]

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)

[beezy]

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

[cherryisripe]

Very nice problem and brilliant explications. Thanks a lot

[malikahashami]

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.

[mom-MA]

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

[epikherolol]

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 +∞

[snowman]

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

[YingchenHe-ks]

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

[KingGisInDaHouse]

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

lim x->-inf sqrt([2x^7+1/4]^2)-2x^7
|2x^7+1/4|-2x^7
|-inf|-(-inf)=inf

[vishalmishra]

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*

[pltrdvl]

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.

[pltrdvl]

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 ...

[sunil.shegaonkar]

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.