Limits of trigonometric functions
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- čas přidán 11. 09. 2024
- This video explains how to find the limits of trigonometric functions.
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Please watch: "Limit of Trigonometric functions at Infinity and non zero values"
• Limit of Trigonometric...
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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.
Oh my God ... Thank u so much for this video 🙏🙏
Thank you! I am glad it was helpful for you!
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.
very helpful video
I am glad, it was helpful. Thanks for visiting my channel, please do share and subscribe to support educational videos.
@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.
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.
Helps a lot
Thanks for visiting my channel. I am glad you like the video. Please share and subscribe to support my channel!
very good
Love you man
Thanks for visiting my channel and supporting the educational videos.
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!!
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
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