Evaluating Limits by Rationalizing | Calculus 1

Поділитися
Вставка
  • Опубліковано 18 вер 2022
  • 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
    Calculus 1 Exercises playlist: • Calculus 1 Exercises
    Calculus 1 playlist: • Calculus 1
    ◉Textbooks I Like◉
    Graph Theory: amzn.to/3JHQtZj
    Real Analysis: amzn.to/3CMdgjI
    Proofs and Set Theory: amzn.to/367VBXP (available for free online)
    Statistics: amzn.to/3tsaEER
    Abstract Algebra: amzn.to/3IjoZaO
    Discrete Math: amzn.to/3qfhoUn
    Number Theory: amzn.to/3JqpOQd
    ★DONATE★
    ◆ Support Wrath of Math on Patreon for early access to new videos and other exclusive benefits: / wrathofmathlessons
    ◆ Donate on PayPal: www.paypal.me/wrathofmath
    Thanks to Petar, dric, Rolf Waefler, Robert Rennie, Barbara Sharrock, Joshua Gray, Karl Kristiansen, Katy, Mohamad Nossier, and Shadow Master for their generous support on Patreon!
    Thanks to Crayon Angel, my favorite musician in the world, who upon my request gave me permission to use his music in my math lessons: crayonangel.bandcamp.com/
    Follow Wrath of Math on...
    ● Instagram: / wrathofmathedu
    ● Facebook: / wrathofmath
    ● Twitter: / wrathofmathedu
    My Math Rap channel: / @mathbars2020

КОМЕНТАРІ • 12

  • @WrathofMath
    @WrathofMath  7 місяців тому

    The ULTIMATE Calculus Christmas Present: ua-cam.com/video/p1s9CufNX60/v-deo.html

  • @jan-willemreens9010
    @jan-willemreens9010 Рік тому +1

    ...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

  • @ftbex9224
    @ftbex9224 Рік тому

    nice.Thanks!

  • @RachaelAnike
    @RachaelAnike 16 днів тому

    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?

  • @infectedswarm3078
    @infectedswarm3078 Рік тому +1

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

    • @WrathofMath
      @WrathofMath  Рік тому

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

  • @JosephMutua-tv7bs
    @JosephMutua-tv7bs 9 місяців тому

    Understood 👍

  • @RuthChikapa-bk8yn
    @RuthChikapa-bk8yn 3 місяці тому

    Thanks sir🥺✨

  • @punditgi
    @punditgi Рік тому +1

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

    • @itzz_just_me2984
      @itzz_just_me2984 Місяць тому

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

  • @VinayakTupkari-je3wn
    @VinayakTupkari-je3wn 2 місяці тому

    Kon kon hindi samjata hai 😂😂😂😂