This annoying problem is testing my limits.

In this video, Ninjas, we discuss how this limit problem can cause frustration since you can't use a quick shortcut like L'Hopital's rule (since the absolute value of sin x isn't differentiable at x = 0)! We need to evaluate this limit by taking a right hand limit and right hand limit due to the absolute value of sin x being present. We'll uncover how this makes the problem more straightforward to solve, and we'll have to use the special limit known as the squeeze limit theorem limit to solve the left hand side limit specifically! I hope you enjoy this video and CRUSH your next exam!

0:00 - Introduction
0:13 - Why L'Hopital's Fails
0:30 - Right Hand Limit
1:51 - Left Hand Limit
3:16 - Squeeze Theorem Limit
4:51 - Simplifying The Limit

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📚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:

Here’s a great study guide so you can CRUSH your AP exam, like a ninja!

This graphing calculator is a beast and never failed me in college:

I loved THIS ruler in college, for engineering classes:

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[NumberNinjaDave]

Make sure to SUBSCRIBE HERE so you don’t miss my tips and tricks for your next exam!

🛍 Want some 🥷🏿 swag? Shop my goodies HERE!


📚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:

Here’s a great study guide so you can CRUSH your AP exam, like a ninja!

This graphing calculator is a beast and never failed me in college:

I loved THIS ruler in college, for engineering classes:

These are my affiliate links. As an Amazon Associate I earn from qualifying purchases.


I use VidIq to help create the best Youtube videos for you! You can sign up here with my affiliate link:
Note that I do make a small commission if you sign up through that link.

[capybara]

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.

[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.

[MASHabibi-dd]

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

[ralvarezb]

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.

[RealQinnMalloryu]

{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 ➖ =sin1.1x^1.1 sinx^1^1 (sinx ➖ 1sinx+1).

[michaelriberdy]

The denominator is not "growing and growing"