This video explains how to find the limits of trigonometric functions. ~-~~-~~~-~~-~ Please watch: "Limit of Trigonometric functions at Infinity and non zero values" • Limit of Trigonometric... ~-~~-~~~-~~-~
It can be achieved like Cos(2x)= Cos(x+x)=cosxcosx-sinxsinx Cos(2x)= cos^2x-sin^2x And cos^2x= 1-sin^2x Therefore Cos2x=1-sin^2x - sin^2x Cos2x=1-2sin^2x Or 1-cos2x=2sin^2x I hope this will help! Thanks for visiting my channel!!
Apply L’Hopital’s rule, it give you lim x->2(sec^2(x-2)/2x) and then plug in x=2. That will give you sec^2(2-2)/2.2 and then you simplify to get the answer 1/4
@@DRMath can i use lim x->2 Tanx/x which equel 1 or it works only when x approaches to zero My teacher didn’t teach me about L’Hopital’s rule yet. I’m new to the derivatives
I'm confused at 23:40 what happens to the denominator (sin^2(x))/x and if 4x(sin^2(2x))/4x turns into ((sin2x)/2x) • ((sin2x)/2x), wouldn't ((sin2x)/2x) • ((sin2x)/2x) turn into (sin^2(2x))/4x^2? Please help.
Dear Blake you are right, i made a mistake there. It suppose to be 4x^2 and x^2 in the denominator. And then both x^2 will cancel out and we will left with only 4. I forgot to put square on x in denominator and numerator. Good observation, keep learning and if you see any other correction please let me know. I make these videos in very short time that I get free after work. Thank you again!!
@@DRMath Hello! I solved it also using identities but I used a different one as yours. I used cos^2(x)-sin^2(x) instead of 2cos^2(x)-1 but I came up with an answer of 2 instead of 4. Can you explain why it doesn't come up with the same answer although I used a double angle identity for cosine? It's making me sad :(
This video explains how to find the limits of Trigonometric functions and plenty of examples are solved to clear the concept.
DR. Math hello
@@ahmedezzat3224 Yes you are right, I made a typo mistake . it suppose to be 4x^2 in the numerator and x^2 in the denominator
some of these questions matched my textbook exactly, so grateful
hello, may i ask what textbook you are talking about? I'm currently busting my butt solving these problems for an upcoming test.
Great video, though I have a problem, if you were asked to find the limit of |cosx|/x as x approaches 0, what would your answer be?
Thanks for visiting my channel, the answer to your question is limit does not exist for this function I believe.
Oh my God ... Thank u so much for this video 🙏🙏
Thank you! I am glad it was helpful for you!
@Harjeet kumar, I have a quick question. Is 1-Cos2x = 2Sin^2x an identity or you made an algebraic manipulation?
It can be achieved like
Cos(2x)= Cos(x+x)=cosxcosx-sinxsinx
Cos(2x)= cos^2x-sin^2x
And cos^2x= 1-sin^2x
Therefore
Cos2x=1-sin^2x - sin^2x
Cos2x=1-2sin^2x
Or
1-cos2x=2sin^2x
I hope this will help!
Thanks for visiting my channel!!
This really helped. Thanks so much.
I will appreciate if you can post more brainstorming questions for us to solve. Thanks once again.
very helpful video
I am glad, it was helpful. Thanks for visiting my channel, please do share and subscribe to support educational videos.
amazing, many thanks
Thank you for visiting my channel.
im lost at 13:16 where did cosx in the denominator came from? also nice video
We wrote tanx as sinx/cosx and this cosx becomes the denominator.
Love you man
Thanks for visiting my channel and supporting the educational videos.
Helps a lot
Thanks for visiting my channel. I am glad you like the video. Please share and subscribe to support my channel!
Find limit tan(x-2)/x^2-4 when x approach 2?
How can i solve this problem
Btw x approach 2 not zero
Apply L’Hopital’s rule, it give you lim x->2(sec^2(x-2)/2x) and then plug in x=2. That will give you sec^2(2-2)/2.2 and then you simplify to get the answer 1/4
I hope this will be helpful
@@DRMath can i use lim x->2 Tanx/x which equel 1 or it works only when x approaches to zero
My teacher didn’t teach me about L’Hopital’s rule yet. I’m new to the derivatives
very good
I'm confused at 23:40 what happens to the denominator (sin^2(x))/x and if 4x(sin^2(2x))/4x turns into ((sin2x)/2x) • ((sin2x)/2x), wouldn't ((sin2x)/2x) • ((sin2x)/2x) turn into (sin^2(2x))/4x^2? Please help.
Dear Blake you are right, i made a mistake there. It suppose to be 4x^2 and x^2 in the denominator. And then both x^2 will cancel out and we will left with only 4. I forgot to put square on x in denominator and numerator. Good observation, keep learning and if you see any other correction please let me know. I make these videos in very short time that I get free after work. Thank you again!!
@@DRMath Hello! I solved it also using identities but I used a different one as yours. I used cos^2(x)-sin^2(x) instead of 2cos^2(x)-1 but I came up with an answer of 2 instead of 4. Can you explain why it doesn't come up with the same answer although I used a double angle identity for cosine? It's making me sad :(
Super
very good vg
Gagan Bawa thank you!!
شكرا
من وين انت ؟
الاردن
Hello
TopThreePlayz hi