FINDING LIMITS ALGEBRAICALLY - CALCULUS

Today, we're learning how to find limits algebraically, when plugging in values for 'x' leads to a scenario where we get 0/0.

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[AnnieVictory]

The one from the exercise where the limit goes to -1 has to be (2*x^2+3*x+1)/(x^2-2*x-3) instead of (2*x^3+3*x+1)/(x^2-2*x-3), right?

[yoinkling]

Thank goodness for youtube. Thank you for this video.

[rosette_renah]

Beautiful video especially for those who are trying to read ahead in this quarantine, like me.
Love this.
Thank you😁😁

[dasimems]

So good, I love it, it really helped me a lot... Thanks

[johnathanbates]

In your last example how exactly did you factor 2x^3 + 3x + 1 to get (2x +1)(x + 1)? If I'm not mistaken you cannot factor this. Perhaps you meant 2x^2 and not 2x^3? Other than that good vid. Thank you for posting.

[theblindbandit_]

I'm just a bit confused at 11:19 because I've never seen such a way of finding the common denominator. Could you explain how it works?

[RetroWes]

A slight correction for the first technique. When you factor out the numerator and cancel one of the factors with the denominator, you are not changing the function.

The domain of a function is a part of its definition. When you cancel out that factor, the technically correct thing to write would be lim x->0 (x+3), x ≠ 2.

[askaaaaaaaaaaaa]

Good video.
Because it is a only channel explain in English ( ch- limits and derivatives)
Please do the video more and more.
It is really helpful video 🙏

[maxmason]

You say in the first example that the curves of x^2+x-6/x-2 and x+3 are not the same and you even stress this point is important. This shook my very mathematical core and I spent longer than I would like to admit trying to work out why the curve had changed but it hasn't. The whole point is that y=x^2+x-6/x-2 is equivalent to y=x+3. The only difference is that y=x+3 is continuous and therefore we can find a value at 2.

[alieukebbeh]

Hello Sir, this is really helpful but there is a little mistake at the last exercise the power of 2x should be 2 not 3, if it's 3 the function Does Not Exist. Thank you!

[bronzedonhandywork]

lim as x approaches -1 (2x^3+3x+1)/(x^2-2x-3) is wrong. Not equal to 1/4. limit does not exist. If it read lim as x approaches -1 would be 1/4

[necatiakpnar]

Thanks for the reminder :) Those methods is the best.

[ysolomon]

the only question I have is, is there a point where we decide that its limit doesn't exist, or do we keep simplifying till we have reached the simplest form?

[laudfx]

the last exercise got me a little bit confuse . the 2xcube is rather 2xsquare.

[Safavster]

I am factoring a problem but I am still left with (x-1) in the denominator which will give me 0. Does this mean I need to use another method?

[rosanarodriguez]

How would you find the limit if an infinity.

[askaaaaaaaaaaaa]

Please do the video about Easy trics in chapter limits and derivatives. ( Ncrt 1st PUC)

[hephzibahakinmade]

I do not know if you will see this comment but I kindly wanted to ask what application you are using and on what device you are using it on?

[GHOST-qxwi]

What happend with the x that was multiplying8 on the 4:11? I mean we cancel out the x^ 2 with the x and it still remains x-8x?

[abdourahmanjallow]

But the upper part is a polynomial not a quadratic equation.(2x+1)(x+1)is not equal to(2x^3+3x+1).
May be it should be (2x^2 instead.)