Fall semester 2020 third class meeting due to covid - 19 we are online learning from NYC CUNY Blackboard Collaborate Ultra. It is not the best template, but having a professor that cannot give simple explanation like Mr Brian Veitch make it worse. Thank you professor Veitch for sharing this top secret information that other professor vowed not to disclosed to student when they took the secret oath of becoming a professor.
At the end of the video, can it be posible for having a point in the same location as the hole? Like for example: f(-2)=-1 & lim x->-2 - ( limit as x approaches negative 2 from the left) or did I do something wrong?
You can have a point in the same location as a hole. It just makes the point a closed dot. The hole is made from the limit, i.e., lim x->-2 f(x) = 4. This gives an open dot at (-2,4). But you can also have a closed dot there if you're also given f(-2) = 4. If this happens then your function is continuous at x=-2. I hope this answers your question.
Do you mean something like lim x->infinity f(x) = 4? If so, this is the definition of a horizontal asymptote. You would draw a dotted horizontal line at y=4. Then sketch your graph so it approaches y=4 as x gets really big.
You would draw the line y = 2x-5. The lim x->4 (2x-5) is asking where would you expect the y-value to be when x->4. If you look at the graph of 2x-5 you can see it would be lim x->4 (2x-5) = 3.
Thank you very much , you are one of the few that did a video on sketching using limits , most other videos do the vice versa only!
I'm having my first exam in Calc 1 next week and you have saved my life !!!! Thank you so much for sharing.
I have a midterm for calc tomorrow.... YOU JUST SAVED ME SO MUCH TIME! You make it look so easy! Thank you!!!!
thank you so much I couldn't figure this out for the life of me for the past week and now I got it in a morning!
Best video on TH-cam considering the subject! Thank you kind sir!
you sir have my gratitude ty
Fall semester 2020 third class meeting due to covid - 19 we are online learning from NYC CUNY Blackboard Collaborate Ultra. It is not the best template, but having a professor that cannot give simple explanation like Mr Brian Veitch make it worse. Thank you professor Veitch for sharing this top secret information that other professor vowed not to disclosed to student when they took the secret oath of becoming a professor.
That escalated quickly.Ty.
beautifully explained.
Your explanations are so clear. Thank you!
I haven't reached this topic in cal yet but you helped me greatly!
Thank u sir for sharing your knowledge this is a big help..😊
you are a saviour. thanks a lot man
Thanks, very clear, concise and helpful.
Taking connecting the dots to a new level but this was really great!!
This was very helpful. Thank you!
thank you so much for this. it really helped i have a quiz tomorrow
This helped so much! Thanks!
great video man
dude, you the best! ;)
thank you so muchhhh!!! i got ittt
Thank you so much with your video~
Thank you Brian!
Appreciated bro
SOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO HELPFUL THANKS YOU SAVED ME
my professor expected us to know how to already, which is stupid because she's the one that taught us the lesson plan for it..
Excellent video
At the end of the video, can it be posible for having a point in the same location as the hole? Like for example: f(-2)=-1 & lim x->-2 - ( limit as x approaches negative 2 from the left) or did I do something wrong?
You can have a point in the same location as a hole. It just makes the point a closed dot.
The hole is made from the limit, i.e., lim x->-2 f(x) = 4. This gives an open dot at (-2,4). But you can also have a closed dot there if you're also given f(-2) = 4. If this happens then your function is continuous at x=-2.
I hope this answers your question.
thanks, such an useful video
Very helpful thank you
really helpful
Thank you soooo much!!!
sir you are great
THIS IS SO HELPFUL
Can I like this more than once?
great video, but what would I do if my limit was approaching infinity rather than equalling it
Do you mean something like lim x->infinity f(x) = 4?
If so, this is the definition of a horizontal asymptote. You would draw a dotted horizontal line at y=4. Then sketch your graph so it approaches y=4 as x gets really big.
Brian Veitch perfect, thanks man
Thankyouuuuu!!!
Thanks so much!
thank you now I can understand
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Thank you
thanks.helped me alot
so helpful!!
Thank you.
thx boss
Thanksss :)
thak you!!!!
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Thank you my lord
FINALLY
I get it!!!!
what if it was Lim x->4 (2x-5) how to draw it ?
You would draw the line y = 2x-5. The lim x->4 (2x-5) is asking where would you expect the y-value to be when x->4. If you look at the graph of 2x-5 you can see it would be lim x->4 (2x-5) = 3.
Thank you
Thank you so much
Thank you so much