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1:46 (sinx-sinx)/x^3 = 0/x^3 = 0. This is equivalent function except at x = 0. Since lim x --> c f(x) = lim x -->c g(x) where g(x) = f(x) for x != c, the limit is 0. No need to substitute in the denominator, it is not 0/0.
What about from the other side? It's not (sinx-sinx), it's |sinx|-sinx. Which means that when you have |sin(-x)|-sin(-x), what you end up with is 2sinx, not 0. A limit only exists if it approaches the same value from both positive and negative.
@@eddiefirstenberg1000 Yes, the limit is -inf from the left side and 0 from the right, which means the limit does not exist. However, the one sided limit from the right exists.
Better than L'Hopital is the developement on Mc Laurin series then try to cancel terms. first to check (I didn't see the video yet), the limit is annoying because Mc Laurin series may not work since |sin(x)| is not derivable, The following identity holds |sin(x)| = sqrt(sin²(x)), you migth take the square of the limit then you'll find the terms like sin(x)|sin(x)| which are derivable near 0. let's see the video If my bet is right.
@MASHabibi-d2d incorrect, but I can see why you think it’s infinity. Remember, formally, a limit talks about values in the neighborhood of the point a and that the function doesn’t need to be defined at f(a) Your homework: what does the value of the fraction equal for any small value of x not equal to 0? If you try 0/0.00001, 0/0.0000000001, etc what do you always get? What value are we approaching for the limit
@@NumberNinjaDaveThe denominator is getting smaller and smaller. The limit in question is of 1/x^2 as x -> 0- x^2 is the denominator, which approaches 0. The reason 1/x^2 approaches infinity is because as x^2 gets smaller, 1/x^2 gets larger. The fraction grows larger and larger, not the denominator.
@ well done! You paid attention to the video and caught my error! I’m glad you’re finding value in my videos Keep the corrections coming. It boosts my videos to help it grow 😉
Make sure to SUBSCRIBE HERE so you don’t miss my tips and tricks for your next exam!
tinyurl.com/numberninjadave
🛍 Want some 🥷🏿 swag? Shop my goodies HERE!
tinyurl.com/numberninjaswag
***********************************************************
📚Helpful stuff and my favorite MUST haves I used in my college courses ⬇
Math and school making you anxious? I totally get it and wrote this book for YOU:
amzn.to/3Y2LWKv
Here’s a great study guide so you can CRUSH your AP exam, like a ninja!
amzn.to/3N5pjPm
This graphing calculator is a beast and never failed me in college:
amzn.to/4eBNeRS
I loved THIS ruler in college, for engineering classes:
amzn.to/4doupRk
These are my affiliate links. As an Amazon Associate I earn from qualifying purchases.
***************************************************************************************
I use VidIq to help create the best RUclips videos for you! You can sign up here with my affiliate link:
vidiq.com/numberninjadave
Note that I do make a small commission if you sign up through that link.
1:46 (sinx-sinx)/x^3 = 0/x^3 = 0. This is equivalent function except at x = 0. Since lim x --> c f(x) = lim x -->c g(x) where g(x) = f(x) for x != c, the limit is 0. No need to substitute in the denominator, it is not 0/0.
What about from the other side? It's not (sinx-sinx), it's |sinx|-sinx. Which means that when you have |sin(-x)|-sin(-x), what you end up with is 2sinx, not 0. A limit only exists if it approaches the same value from both positive and negative.
@@eddiefirstenberg1000 Yes, the limit is -inf from the left side and 0 from the right, which means the limit does not exist. However, the one sided limit from the right exists.
Better than L'Hopital is the developement on Mc Laurin series then try to cancel terms.
first to check (I didn't see the video yet), the limit is annoying because Mc Laurin series may not work since |sin(x)| is not derivable, The following identity holds |sin(x)| = sqrt(sin²(x)), you migth take the square of the limit then you'll find the terms like sin(x)|sin(x)| which are derivable near 0.
let's see the video If my bet is right.
You were on the right track! Watch until the end to see
The limit of the first part is wrong, the positive answer is infinite, thank you
@MASHabibi-d2d incorrect, but I can see why you think it’s infinity.
Remember, formally, a limit talks about values in the neighborhood of the point a and that the function doesn’t need to be defined at f(a)
Your homework: what does the value of the fraction equal for any small value of x not equal to 0? If you try 0/0.00001, 0/0.0000000001, etc what do you always get? What value are we approaching for the limit
Now it's done...thank you
@@MASHabibi-d2d mabrouk
{sinx+sinx ➖} ➖ sinx/x^3= sinx^2 ➖ (sinx)^2/x^3={sinx^2 ➖ sinx^2}/x^3 =sin{x^0+x^0 ➖} sin{x^0+x^0 ➖ }={sinx^1+sinx^1}/x^3=sin^2x^2/x^3 =sin1.1x^1.1 sinx^1^1 (sinx ➖ 1sinx+1).
The denominator is not "growing and growing"
Are you sure 🤔
@@NumberNinjaDaveThe denominator is getting smaller and smaller.
The limit in question is of 1/x^2 as x -> 0-
x^2 is the denominator, which approaches 0.
The reason 1/x^2 approaches infinity is because as x^2 gets smaller, 1/x^2 gets larger.
The fraction grows larger and larger, not the denominator.
@ well done! You paid attention to the video and caught my error! I’m glad you’re finding value in my videos
Keep the corrections coming. It boosts my videos to help it grow 😉