I really like how you explained it! But I almost didn't click on the video because of the thumbnail. You should try some picture with you as it would look more legit. Just my take :)
Isn't it easier to just use mclaurin here so that you get X-X-1/3!(x^3) / x^3 then you get x^3/6 / x^3 and from here you can multiply both for 1/x^3 and thus get the answer? or am I doing anything wrong?
![Limits Using Triple Angle Identity[Calculus]](http://i.ytimg.com/vi/eTJ3CyV2pq4/mqdefault.jpg)
![Limits Using Triple Angle Identity[Calculus]](/img/tr.png)
Nicely done sir. 👍
thank you so much, I haven’t seen better explanation before
Thank you so much! Respect!
Sir Thank you so much, tomorrow is my maths paper and this video really helped me..
Godspeed!
Thank you so much for the help!
I really like how you explained it! But I almost didn't click on the video because of the thumbnail. You should try some picture with you as it would look more legit. Just my take :)
Thank you for your video.....sr
very good explain thanksss
thank you
Hello sir I been wishing to chat with u thanks for the video
Isn't it easier to just use mclaurin here so that you get X-X-1/3!(x^3) / x^3 then you get x^3/6 / x^3 and from here you can multiply both for 1/x^3 and thus get the answer? or am I doing anything wrong?
Squeeze theorem could help but you should choose inequalities correctly
I prefer triple angle approach to this limit
Could you give the solution please