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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[bryantkensotero]

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

[jan-willemreens]

...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

[ishtiaquekhan]

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

[jazou]

Thanks for this! I thought conjugate was the only way to solve this kind of problem. You earned a subscriber! <3

[justinpark]

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.

[pianoplayerable]

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

[evangelistajaymar]

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

[jee-lysis]

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

[jinmin]

What if the radicals is in denominator?

[Idontcaretomakearealusername]

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