This annoying problem is testing my limits.

[@NumberNinjaDave]

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[@capybara341]

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.

[@eddiefirstenberg1000]

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.

[@capybara341]

@@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.

[@robertlunderwood]

For the homework, there's a difference between the limit of a function going to zero and the function actually being zero. The numerator wasn't going to zero; it was zero. The denominator would've gotten smaller and smaller, but since the numerator was 0, it didn't matter. Apply L'hopital all you want.

[@NumberNinjaDave]

Good observation

[@MASHabibi-d2d]

The limit of the first part is wrong, the positive answer is infinite, thank you

[@NumberNinjaDave]

@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

[@MASHabibi-d2d]

Now it's done...thank you

[@NumberNinjaDave]

@@MASHabibi-d2d mabrouk

[@ralvarezb78]

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.

[@NumberNinjaDave]

You were on the right track! Watch until the end to see

[@RealQinnMalloryu4]

{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).

[@michaelriberdy475]

The denominator is not "growing and growing"

[@NumberNinjaDave]

Are you sure 🤔

[@ethannguyen2754]

@@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.

[@NumberNinjaDave]

@ 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 😉