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!

[subbaraooruganti]

Excellent way to handle absolute value functions

[m.h.]

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.

[CalculusIsFun]

I thought of it like this. Thinking of the top one like a piece wise function, it can be interpreted as two linear functions. The bottom one is a cube root. The top two are growing faster than the bottom and both go to zero, and the bottom also goes to zero but at a slower rate. This the top takes precedence and drives the limit to zero.

[ГригорийКузярин-тъ]

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

[Grecks]

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.

[pojuantsalo]

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.

[Christian_Martel]

5:05 I beg to differ here. The first piece is true when x >= 3 or x<= -3, with the second piece true when -3 < x < +3

[ruchirgupta]

You said |x^2 - 9| = -(x-3)(x+3) when x<3. But what if x = -7? If x = -7, -(x-3)(x+3) = -40, which is not a positive value. Actually you missed that |x+3| = x+3 for (x>= -3) and |x+3| = -x-3 for (x<-3).

[annacerbara]

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. 😊

[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.

[diogochadudmilagres]

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!

[yasinforughi-bz]

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

[dieuwer]

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.

[Targetst]

Sir please solve the jee advanced math paper 2023
Paper1 and Paper2
It would be very helpful and indeed an exciting journey
Love from India❤❤

[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.

[jakehobrath]

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

[mimi-kulb]

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

[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.

[srisaishravan]

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!