Evaluating Limits by Rationalizing | Calculus 1

We find limits by rationalizing the numerator (or rationalizing the denominator, it works out very much the same). We'll do two examples, we'll find the limit of (sqrt(x-1)-1)/x as x approaches 0 and the limit of (sqrt(x+1)-2)/(x-3) as x approaches 3. #calculus1 #calculus

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[jan-willemreens]

...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

[infectedswarm]

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

[RachaelAnike]

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?

[punditgi]

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