The ULTIMATE Calculus Christmas Present: kzread.info/dash/bejne/omWnm6WulrG-ZpM.html

...Good day to you, Thank you for your clear presentation. After watching I want to share with you an alternative way to solve the same two indeterminate limits: 1) For your limit: lim(x-->0)((sqrt(x+1) - 1)/x) I first start by rewriting the denominator x as follows: x = (x + 1) - 1, then I treat this expression as a difference of squares: x = (x + 1) - 1 = (sqrt(x + 1) - 1)(sqrt(x + 1) + 1), and finally after replacing the denominator x in your original limit by this last factored expression, we can cancel the common factors of top and bottom, resulting in the solvable limit: lim(x-->0)(1/(sqrt(x+1) + 1)) = 1/2, and 2) For lim(x-->3)((sqrt(x+1) - 2)/(x - 3)), first rewrite the denominator x - 3 as follows: x - 3 = (x + 1) - 4, again think of difference of squares = (sqrt(x+1) - 2)(sqrt(x+1) +2), resulting after cancelling common factors of top and bottom in the following solvable limit: lim(x-->3)(1/(sqrt(x+1) + 2)) = 1/(2 + 2) = 1/4; the same outcome! I hope you appreciate this way of solving too. Finally, this was my Wrath ON Math (lol)... Thank you for your math efforts, and take care, Jan-W

Please when do we know if we're to rationalize the numerator or denominator....and hoe do we know when to use the differentiation method or rationalization method in an equation?

nice.Thanks!

Understood 👍

thank you so much for this wonderful video.. You nicely explained the concept..and plz post more videos about calculus.

@WrathofMath

Жыл бұрын

Thank you! Any specific requests for what calculus videos you want to see next?

Thanks sir🥺✨

@WrathofMath

4 ай бұрын

Most welcome!

I thought real square roots had both a positive and a negative solution. Why ignore the negative root?

@itzz_just_me2984

2 ай бұрын

because its an equation not an expression, you write it as is, not with a plus/minus

Kon kon hindi samjata hai 😂😂😂😂