Find the limit

In this video, i used the piecewise definition of absolute value to simplify the function, taking one-sided limits was the next important step.

[@AndreasHontzia]

I first extended the absolute function to the denominator, which was an oversight by me, as it is the 3rd root, which means it can be negative. If it would be an even root, it would be possible, as the sign does not change. But it provided me with a glimpse of turning the fraction into a multiplication. So I knew I was on the right track. Thank you for this very nice solution!

[@AndreasHontzia]

Maybe I found a different argument. So let's extend the absolute value to the denominator. For this to work, we will have to compensate for the sign (+/-) of the denominator. The sign functions sgn is defined as 1, if the argument is positive, -1 if the argument is negative and 0 if the argument is 0. This will lead to: sgn((x-3)^(1/3)) * | (x²-9) / (x-3)^(1/3) | You can now proceed like in the video to get: sgn((x-3)^(1/3)) * | (x-3)^(2/3)*(x+3) | Applying the limit will result in 0 * 0 * 6 = 0.

[@Grecks75]

Answer: The limit is 0. The function term can be written as a product: |x + 3| * sgn(x - 3) * (x - 3)^(2/3) where the first two factors are bounded and the last one converges to 0 as x approaches 3. Therefore, the entire product must converge to 0.

[@m.h.6470]

Solution: First, calculate limit 3⁺, so x = 3 + ε |x² - 9| / ³√(x - 3) = |(3 + ε)² - 9| / ³√(3 + ε - 3) = |9 + 6ε + ε² - 9| / ³√ε = |6ε + ε²| / ³√ε = (6 + ε) * ε/³√ε = (6 + ε) * (³√ε)³ / ³√ε = (6 + ε) * (³√ε)² which is just above 0, so the limit 3⁺ is 0 Then calculate limit 3⁻, so x = 3 - ε |x² - 9| / ³√(x - 3) = ³√( |x² - 9|³ / (x - 3) ) = ³√( |(3 - ε)² - 9|³ / (3 - ε - 3) ) = ³√( |9 - 6ε + ε² - 9|³ / -ε ) = ³√( |ε² - 6ε|³ / -ε ) since ε is a tiny positive value and x² - 6x is < 0 for values between 0 and 1, we know, that |ε² - 6ε| is a negative value = ³√( -(ε² - 6ε)³ / -ε ) = ³√( ((ε - 6) * -ε)³ / -ε ) = ³√( (ε - 6)³ * (-ε)³ / -ε ) = (ε - 6) * ³√( (-ε)³ / -ε ) since the exponent of (-ε)³ is odd, we can write it as -ε³ = (ε - 6) * ³√( -ε³ / -ε ) = (ε - 6) * ³√(ε²) which is just below 0, so the limit 3⁻ is 0 Therefore the limit of 3 for |x² - 9| / ³√(x - 3) is 0.

[@ruchirgupta610]

You said |x^2 - 9| = -(x-3)(x+3) when x= -3) and |x+3| = -x-3 for (x

[@srisaishravan5512]

Yes

[@BP-gn2cl]

Logical point

[@msgamerz2998]

@@ruchirgupta610 same thought bro

[@Merched45]

I think he just didn't bother because in the problem x approaches 3, so these numbers don't really matter. But I'm not sure

[@SalimHadid-h6p]

We are evaluating the limit at 3, so we only care about what happens when x is close to 3 (but never equals 3). if you want to know what happens at x=-7, just replace it in the original function.

[@subbaraooruganti]

Excellent way to handle absolute value functions

[@ГригорийКузярин-т1ъ]

I did it the other way: I replaced √x-3 with n so x is n³+3 and the limit is |n⁶+6n³+9-9|/n with n approaching 0. If we take out n² and divide it by n what is left is n*|n⁴+6n|=0*|0|=0

[@DanMusceac]

It's a clever way to solve this problem.

[@annacerbara4257]

It can be said that (x-3) above is an infinitesimal of higher order than (x-3)^1/3 below therefore it prevails and therefore the function tends to zero. 😊

[@pojuantsalo3475]

After dividing the original limit with absolute value into two limits from the left and from the right I applied L'Hôpital's rule. Both one-sided limits turned out 0 meaning the limit exists and is 0.

[@diogochadudmilagres4533]

According to the graph, there's an inflection point in x = - 3. Coul'd you calculate the limit in this point for another video? Thanks!

[@seanmsgjuan]

I believe you also should have defined the absolute value in the piece-wise function in relation to -3. -4, for example, yields a negative result as shown. I don't think it's as relevant given the limit is approaching 3, but it felt incomplete without it.

[@ChristopherBitti]

|x^2 - 9| either takes the value x^2 - 9 or 9 - x^2 so if you can prove the limit for both (x^2 - 9)/(x - 3)^(1/3) and (9 - x^2)/(x - 3)^(1/3) then we are done. It is easy to show both these limit to 0 at x = 3 using L'Hopital's rule.

[@yasinforughi-b1z]

In one of your videos on solving the improper integral of ln^2(x)/x^2+1 You used a Fourier series expansion but I didn't really understand it Please make a video on that part

[@srisaishravan5512]

Day 4 of asking newton sir to start making videos on integrals. Edit: I typed before seeing the video. Thank you sir for heeding to my wishes and making videos on calculus! But i would still like to see vids on integrals. Thanks!

[@henrymarkson3758]

Why does every math channel have to do integrals? There is plenty of integration content on KZbin, it's getting boring

[@yasinforughi-b1z]

Actually he has made LOTS of videos on integrals if you check his playlists but not many on geometry

[@devcoolkol]

Go watch blackpenredpen

[@PrimeNewtons]

I believe I have done more videos on integrals than most other topics on this channel. See my calculus playlist.

[@Grecks75]

I'd much rather stick with problem solving in general, it's more interesting and creative. If that includes an integral, fine. A pure integral once in a while? Fine. But there is so much more than integrals. Resolving integrals is mostly about technique and methods, it gets boring soon. I could ask Prime Newtons to do more videos on counting (sic, no joke!) but I won't. 😃 It's also about technique, yet more interesting and insightful than integrals. I'm satisfied with his choice, though.

[@dieuwer5370]

This trick only works because of the (odd) cube-root in the denominator. For even roots in the denominator, it becomes imaginary. Then you cannot simply divide the first term of the nominator by the denominator.

[@Grecks75]

Most likely, this problem is from Real Analysis. In case of a square root in the denominator, the domain of the function would be restricted to x > 3. Dividing works the same as before, and the limit would still be 0, similar to a one-sided limit approaching from above.

[@emanahmed745]

Amazing ❤

[@DanMusceac]

You did a mistake when you said that / x+3 /is the same no master the value of x.If x is less than -3 than you must written 3-x.

[@Viki13]

He said that because x approaches 3 so abs(x+3) can just be set equal to x+3 since the result will definitely be positive

[@MichaelGrantPhD]

@@DanMusceac you only have to be concerned about the neighborhood around +3, due to the limit. He definitely could have been clearer about that, but he wasn't wrong.

[@msgamerz2998]

I want geometry!!!!!!!!. Day 1 of my protest.

[@koenth2359]

Maybe throw some soup on a famous painting if you wanna be heard

[@surendrakverma555]

Thanks Sir 🙏🙏🙏🙏

[@BBG-Bob]

Live this channal

[@mimi-ku3lb]

Sorry but as the third root is definiert only for nonnegatives, x-3 has to be positive so no 3- approach from left side

[@PrimeNewtons]

Left side does not mean negative. It means les than

[@Grecks75]

It's a matter of definition. In some text books the real n-th root function is only defined for positive real x, but you can also uniquely extend that definition to zero and, in case of odd n, to negative x. With the latter definition you must consider both one-sided limits for the proper limit.

[@amitanshsrivastav9640]

A sphere of radius 5 cm is placed inside a beaker filled with water. The height of the beaker is 50 cm and base is 60 cm. The density of the water is 1000kg/m³ and atmospheric pressure is 101 kilo pascals. What will be the pressure experienced by the top point of the sphere ?(Take gravity=10m/s²) Options:- 1) 105 kilo pascals 2)100 kilo pascals 3)110 kilo pascals 4)115 kilo pascals This is from the National science Olympiad, India, class 8

[@francescobedinijacobini]

You can find limit by comparing the "orders of infinitesimal" here denoted as "ord". An order of infinity would be denoted as Ord. The numerator is ord(|x^2 -9| = 2 while the denominator is ord(CUBRT(x - 3)) = 1/3. If you compare the two orders, the numerator approaches zero faster than the denominator. Thus the limit is 0.

[@peshepard412]

A cusp at x = 3

[@pologuy1637]

Hello sir, can you please solve the simultaneous Logarithmic equations, that I sent you via email. Thanks.

[@NamuncuraSalcedo]

Thanks for the breakdown! Just a quick off-topic question: My OKX wallet holds some USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). How can I transfer them to Binance?

[@jakehobrath7721]

Ah the factoring was clever, I used l’hopital and found the limit approaches zero on both sides

[@SSahoo-jk7vc]

Hi