Evaluating limits by Conjugate Method - Limits and Radicals - Calculus

2022 ж. 23 Нау.
62 224 Рет қаралды

In this video we will learn how we can find limits of functions by the conjugate method.
To solve certain limits, you need the conjugate multiplication technique. When substitution doesn’t work in the original limit then you can use conjugate multiplication to manipulate the function until substitution does work. This method works for limits involving radicals.
In mathematics, the limit of a function is a fundamental concept in calculus and analysis concerning the behavior of that function near a particular input.
Formal definitions, first devised in the early 19th century, are given below. Informally, a function f assigns an output f(x) to every input x. We say that the function has a limit L at an input p, if f(x) gets closer and closer to L as x moves closer and closer to p. More specifically, when f is applied to any input sufficiently close to p, the output value is forced arbitrarily close to L. On the other hand, if some inputs very close to p are taken to outputs that stay a fixed distance apart, then we say the limit does not exist.
The notion of a limit has many applications in modern calculus. In particular, the many definitions of continuity employ the concept of limit: roughly, a function is continuous if all of its limits agree with the values of the function. The concept of limit also appears in the definition of the derivative: in the calculus of one variable, this is the limiting value of the slope of secant lines to the graph of a function.

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  • Thank you so much sir... ❤

    @nishusahu3681@nishusahu36814 ай бұрын
  • you made me learned enough sir long live sir

    @allandizon5601@allandizon56013 ай бұрын
  • This is excellent! Where can I find a gazillion conjugation problems to work out so I can become confident and proficient in that method??

    @elmoreglidingclub3030@elmoreglidingclub30303 ай бұрын
  • Good work, Does limit solve undefined form or indeterminate form type of problems?

    @jonykhan4395@jonykhan43954 ай бұрын
  • Nice 👍

    @arslanmoghal5472@arslanmoghal54723 ай бұрын
  • thankyou sirrr

    @jayfrancisabraham7921@jayfrancisabraham79213 ай бұрын
  • very nice and thorough method of explanation, thank you ^_^

    @adamk5701@adamk57016 ай бұрын
  • Bro u r an absolute G, bc of u I now only have an 80% chance of failing my calc exam rather than 100% Update: I got a 92%, this geezer is the real deal

    @GreaseMonkey33@GreaseMonkey333 ай бұрын
  • Can someone pls tell me the method or the process of how he found the common denominator at 12:35

    @corazonnn_@corazonnn_3 ай бұрын
    • The common denominator of the denominators x and √1+x is just their product, since they have no common factors. This is the same situation if you wanted the common denominator of 1/5 and 1/7, which is the product 5*7.

      @xoppa09@xoppa093 ай бұрын
  • Very good method

    @bigfoot4034@bigfoot40348 ай бұрын
  • Good vedio!!!

    @SisayInjigu-vm7gq@SisayInjigu-vm7gq Жыл бұрын
  • thx for this video

    @illusoriokarlvincentv.2019@illusoriokarlvincentv.2019 Жыл бұрын
  • Good mathes

    @SabaaOromo@SabaaOromo25 күн бұрын
  • Can i ask why in the last question, why is it not possible to combine (√x+1) and (1+√x+1)?

    @meliodassama988@meliodassama9888 ай бұрын
    • if you talk about denominator, it makes it more complicated

      @calculus997@calculus9978 ай бұрын
  • While this is good the probem is that what are the odds that the numerator will have a common factor as the denominator? So many special cases when some general rule would be nice or graphically representing the expression and estimating the answer by observation. Let AI deal with the problem.

    @slottibarfast5402@slottibarfast5402Ай бұрын
  • ... Good day, An alternative way to solve the given indeterminate lim(x-->2)((SQRT(4x + 1) - 3)/(x - 2)) ... Rewrite the denominator (x - 2) as follows: x - 2 = (4x + 1) - 9 by first multiplying numerator and denominator by 4, thus: lim(x-->2)(4)((SQRT(4x + 1) - 3)/((4x + 1) - 9)) ... Treat the new denominator (4x + 1) - 9 as a difference of two squares: (SQRT(4x + 1) - 3)(SQRT(4x + 1) + 3) and finally cancell the common factor (SQRT(4x + 1) - 3) of numerator and denominator to obtain the solvable limit: (4)lim(x-->2)(1/((SQRT(4x + 1) + 3)) = (4)1/(3 + 3) = 4/6 = 2/3 ... I hope you appreciate this way too ... Thank you and take care, Jan-W

    @jan-willemreens9010@jan-willemreens9010 Жыл бұрын
    • Yes, it works, but unusual method

      @calculus997@calculus997 Жыл бұрын
  • why is 2/√1+√1 = 2/2 ? shouldn't it be 2/√2? please enlighten me

    @judgemanas9854@judgemanas98546 ай бұрын
    • No, 2/(1+1)=2/2=1

      @calculus997@calculus9976 ай бұрын
  • Why are we taking conjugate of numerator and not denominator

    @AbdulSalam-ss4eu@AbdulSalam-ss4eu8 ай бұрын
    • 8:44 Watch this

      @itsdokuro610@itsdokuro6105 ай бұрын
    • Yes oo man good question

      @abdallahneymar8615@abdallahneymar86154 ай бұрын
    • What are you trying to get rid of?

      @solomonikeoha9074@solomonikeoha90743 ай бұрын
    • We don't always use denominator, we use the term that has square roots to cancel them out

      @jvpro_rblx2955@jvpro_rblx29553 ай бұрын
  • How do you get 1-rad1+x for the numerator, I don’t get it

    @jeremiahgallegos2111@jeremiahgallegos21117 ай бұрын
    • first we did a common denominator, and so the numerator is ...

      @calculus997@calculus9977 ай бұрын
    • Yeah me too

      @itsmemario1298@itsmemario12984 ай бұрын
    • im also confused on that last part on how you got the 1-rad1+x/xrad1+x @@calculus997

      @manuelissiahrito6917@manuelissiahrito69173 ай бұрын
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