Evaluating limits by Conjugate Method - Limits and Radicals - Calculus
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- เผยแพร่เมื่อ 23 มี.ค. 2022
- 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.
3:41 thank you for this method!
Thank you so much sir... ❤
you made me learned enough sir
long live sir
This is excellent! Where can I find a gazillion conjugation problems to work out so I can become confident and proficient in that method??
Good work, Does limit solve undefined form or indeterminate form type of problems?
Nice 👍
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
thankyou sirrr
very nice and thorough method of explanation, thank you ^_^
Can someone pls tell me the method or the process of how he found the common denominator at 12:35
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.
Good vedio!!!
Very good method
thx for this video
Can i ask why in the last question, why is it not possible to combine (√x+1) and (1+√x+1)?
if you talk about denominator, it makes it more complicated
Good mathes
why is 2/√1+√1 = 2/2 ? shouldn't it be 2/√2? please enlighten me
No,
2/(1+1)=2/2=1
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.
Why are we taking conjugate of numerator and not denominator
8:44
Watch this
Yes oo man good question
What are you trying to get rid of?
We don't always use denominator, we use the term that has square roots to cancel them out
... 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
Yes, it works, but unusual method
How do you get 1-rad1+x for the numerator, I don’t get it
first we did a common denominator, and so the numerator is ...
Yeah me too
im also confused on that last part on how you got the 1-rad1+x/xrad1+x
@@calculus997