Limit with a Cube Root - Factoring is the Way to Go! | Limits | Calculus | Glass of Numbers

In this video, we look at a limit with a cube root.
No, don't multiply the numerator and denominator by the conjugate of the expression.
It doesn't work. Instead, we factor x - 64. How do we do it? Watch this video for the details!
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[@k25p16]

Thank you man!

[@GlassofNumbers]

Glad you like this video! Thank you! Please share my videos to others who may need this!

[@RonaldNaibei]

Perfectly presented thanks 🙏🙏

[@GlassofNumbers]

Thank you! Please help me share this video with others 😁👍

[@vibe_of_ayax]

Thank you very much ☺️🙏🏻

[@GlassofNumbers]

Thank you for your comment! Glad it helps! Please share my videos to your friends!!

[@bryantkensotero4155]

thanks man, I almost gave up solving this exact same problem but the radical is in the denominator

[@GlassofNumbers]

Good you didn't give up! It is a difficult one! Same idea when the radical is in the denominator.

[@ahmedsherif1359]

Thanks man, Appreciate it

[@GlassofNumbers]

Glad you like this 😁 please share my video to others!!

[@christianvillacorte6966]

thank you!

[@GlassofNumbers]

Glad you like this video! Please share it with others 😁

[@jan-willemreens9010]

...Good day Wilson, You showed a great alternative way to solve your indeterminate limit by applying the difference of two cubes. If you find the time, let me ask you in return to solve the next 2 indeterminate (0/0) limits also without using the conjugate method: 1) lim(x-->4)((sqrt(x + 5) - 3)/(x - 4)) and 2) lim(x-->7)((2 - sqrt(x - 3))/(x^2 - 49))... In case you found some spare time trying to solve them, I wish you good luck! Hope to hear from you... Thank you for your math efforts and take care, Jan-W

[@GlassofNumbers]

Good idea for video! There are a few more ways (in addition to conjugate method) to find the limit for (1) and (2). I will make a video with different ways 😁👍 Thank you!!

[@jazou388]

Thanks for this! I thought conjugate was the only way to solve this kind of problem. You earned a subscriber!

[@GlassofNumbers]

I am glad that you like this video! Thank you! Yes, the conjugate technique only works well for square roots.

[@adobo4033]

Thank You!

[@GlassofNumbers]

Thank you for your comment! I am glad this video helps! Please subscribe to my channel and share my videos!

[@evangelistajaymar8416]

thanks, it really helps me for my activity, Godbless :D

[@GlassofNumbers]

Good to hear it helps! Thank you! What activity do you do?

[@ishtiaquekhan1666]

thank you so much but what if the denominator is just one variable such as x ?

[@ishtiaquekhan1666]

how could u solve that then? because there is only A and no B in terms of A^3-B^3

[@GlassofNumbers]

Then you can find the limit by direct substitution. Note that if it's just x in the denominator, then it's not an indeterminate form of 0/0 anymore.

[@mariel98210]

@@GlassofNumbers what if the denominator is only x, and the limit is x --> 0 so it will still be an indeterminate form, how to solve that?

[@justinpark939]

How about substituting the variable x^1/3 for u and thus, as x -- > 64, u --> 4 thus we have (u-4)/(u^3-64). I think there was a diff of cubes formula but I don't know it so I just used synthetic division, knowing by factor remainder theorem that u-4 was a factor. Thus, by canceling u-4 in the numerator and denominator, I got 1/(u^2+4u+16), let it approach 4 and get 1/48. I just did the substitution just to make my work more readable for me.

[@GlassofNumbers]

That also works! Good job! 😁

[@justinpark939]

@@GlassofNumbers thanks

[@pianoplayer123able]

If I use de l' Hospital I get the same result. But solving it this way is far more interesting and exciting.

[@GlassofNumbers]

I agree!

[@Idontcaretomakearealusername]

Thank you I was about to kill something wolfram alpha wasn’t even able to explain it to me

[@GlassofNumbers]

Please share my video with others! Thank you!

[@jaysoncagustin]

4^3√64
How doest it equal to 16?

[@GlassofNumbers]

Cube root of 64 is 4. We then multiply by 4, we get 16.

[@jee-lysis233]

great man , you are lim x-0 1/x . thanks for the trick///

[@GlassofNumbers]

Thank you! Glad my video helps!! Please share my video to others 😁

[@Berrybooo]

what about using conjugate

[@GlassofNumbers]

You meant square root conjugate?

[@Berrybooo]

@@GlassofNumbers yes

[@jinmin3471]

What if the radicals is in denominator?

[@GlassofNumbers]

If the radical is in the denominator, then you will still do it the same way as in this video

[@giuseppemalaguti435]

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