Limits at Infinity & Horizontal Asymptotes
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- เผยแพร่เมื่อ 26 ก.ย. 2024
- This calculus video tutorial explains how to evaluate limits at infinity and how it relates to the horizontal asymptote of a function. Examples include rational functions, radical functions, inverse trigonometric functions and exponential functions. This video contains plenty of practice problems on evaluating limits at infinity analytically with radicals, square roots, and trig functions.
Introduction to Limits:
• Calculus 1 - Introduct...
How To Evaluate Limits From a Graph:
• How To Evaluate Limits...
Evaluating Limits By Factoring:
• Evaluating Limits By F...
Limits of Rational Functions :
• Limits of Rational Fun...
Limits of Radical Functions:
• How To Evaluate Limits...
Limits of Trigonometric Functions:
• Limits of Trigonometri...
___________________________________
How To Find The Limit at Infinity:
• How To Find The Limit ...
Infinite Limits + Vertical Asymptotes:
• Infinite Limits and Ve...
Limits at Infinity With Radicals:
• Limits at Infinity Wit...
Limits of Absolute Value Functions:
• Limits and Absolute Value
Limits of Composite Functions:
• How To Find The Limit ...
The Squeeze Theorem:
• Squeeze Theorem
____________________________________
Limits and Logarithms:
• Limits and Logarithms
Limits of Exponential Functions:
• Limits of Exponential ...
Piecewise Functions - Limits:
• Piecewise Functions - ...
3 Step Continuity Test:
• 3 Step Continuity Test...
Continuity and Differentiability:
• Continuity and Differe...
Limits - Test Review:
• One Sided Limits, Grap...
_____________________________________
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections
Limits - Free Formula Sheet:
bit.ly/3T3dD2X
Limits - Free Formula Sheet: bit.ly/3T3dD2X
Final Exams and Video Playlists: www.video-tutor.net/
Math and Science Tutors: bit.ly/Find-a-Tutor
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Can someone please explain why at 10:13 he multiplied both top and bottom by 1/x instead of 1/x²
@The Organic Chemistry Tutor
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My calculus teacher dint even teach how to find the limit of infinity
@@alikhidzam3749 you serious? That should be one of the first things you learn in calc 1
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This was helpful as always, yet it is still challenging to me a little bit because there's no consistency in the procedures so that we can use the same approach to every rational function.
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G 191 11:38
note 5:25 any time the function is bottom heavy with infinity limits the functions is equal to zero
note 5:25 any time degree is equal the ans is coefficient ratio
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16:14 Why did he multiply the radical in the denominator by 1/x^2, but the rest of the denominator by 1/x? I'm guessing it's because in the end, you're taking the limits individually, but I don't get the rule
"Inside a radicle, we need to multiply by 1/x^2" is what he says.
Because the 1/x is going under the radicle, its supposed to turn into
1/x^2.
Hope this helps!
The x doubles inside the sqr rt! Omg thank goodness I was so confused thanks!!!!
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may i ask on the second to the last problem, why do we need to multiply
1/x on both sides?
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At 11:35, since x goes to infinity, I think that we have to consider both case: (1) x goes to positive infinity and (2) x goes to negative infinity. In case (1), the answer is 2 (positive 2), but in case (2) the answer is -2 (negative 2) . In this problem, the method divide both numerator and denominator by 1/x is not a good choice. Instead, divide both numerator and denominator by 1/abs(x) (abs(x): absolute value of x) would be more generally correct. In case (1), 1/abs(x) equals 1/x (positive 1/x) and in case (2), 1/abs(x) = 1/-x (negative 1/x).
at 16:06 shouldntthe 9 be multiplied by 1/x^2 instead of 1/x because x^2/x is still x???? im confused pls help
ok im a year late but it turns out to be 9 because to be able to multiply √(9x^2) by (1/x) to cancel out the highest exponent, you need to get into the radical. So what he did without explaining was he converted (1/x) into (1/√x^2) bc they're the same, so √(9x^2)/(1/x) = √(9x^2)/√(1/x^2) which makes it √9. -bored college student
"The exponent doubles once you move it inside" 19: 21 I bet many of you are confused but hopefully this one sentence help just like it helped me lol
thank youuuu I was confused on that
bruh i was hella confused about this before i saw this video
You made limits so so so so so so so so soooooo much easier. Thank you. For realsies.
One short cut rule you didn't seem to cover was when the x value is greater in the numerator (ex: 3x^4-10/6x^2+20) the limit will be +/- infinity since the top is outgrowing the bottom.
why do we have to multiply the fraction by 1/x😢
to make sure that the x become a denominator with a value of 0
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I haven’t finished the video but I’ve been studying other calc sections for my quiz tomorrow and you just apply the power rule to top and bottom and you get the answer, so all constants = 0, for the constants that have a variable you multiply it by 1 and divide and get the answer. hopefully this helps someone in the future!
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12:28 HOW DO YOU DECIDE WHICH FRACTION YOU SHOULD MULTIPLY WITH
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The answer at 9:20 really helped me out. Thanks a lot!
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In 11:51 why are we multiplying by 1/x³ instead of 1/x since this question is bottom heavy?
Does anyone know?
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At 11:35 the answer should be -2 because when x enters inside the square root it becomes x^2 and you have to multiply the limit by a negative sign.
What you are saying is true when x -> - infinity, not when x -> infinity
quick question, for the problem w (square root of 9x^2+x) - 3x, why couldnt we plug in infinity right away? x^2 is the biggest term, so wouldn't it determine the behavior of the limit and cause it to tend towards infinity?
Can you.please show these carrying the proper Lim notation through the whole problem? Even with the "work" shown, it skips steps or notation which we need to show on every step.
15:50 dude, when did 9x^2 divided by x became 9??
since it is inside a root, you have to divide by 1/x^2 and not 1/x
@@Jacobk2403 thanks boss I understand ur point clearly
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this is so much easier I honestly wish they taught us in precalc
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15:36 why did he multiply the top and the bottom with 1/x instead of 1/x^2 since the highest degree is 2?
same question
Because he got rid of it in the numerator and in the denominator it’s inside a square root. Therefore, x is the highest
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Sir please solve the question.
Evaluate lim n tends to infinite x to the power n divided by e to the power x
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I would like to ask 15:54 ? why is it like 1/x when the numerator is only 1? Thank you
same here, I'm confused at that part, did u figure it out?
thank you sir
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@4:00 why are we multiplying the top and the bottom by 1/x”
x was at the denominator. It was first 4/x then simply 4/1 multiplied by 1/x. In short he separated it. I hope this answer is helpful and sorry I replied 5 months later.
@@champagnestainedbirkin what do you mean "first 4/x then simply 4/1 multiplied by 1/x. In short he separated it. " where did 4/x came from and 4/1?? anyways..whats the purpose of him multiplying (1/x) at the top and bottom?
I think he does it to show WHY the limit of (8/3x+4) is equal to zero. Since he was already explaining why the limit of (1/x) is always zero, he made both the numerator and denominator in terms of (1/x) to prove his point. So you don't need to do it if you already get how and why the limit of the function is equal to zero, but he does it to explain it more thoroughly.
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When you multiplied 1/x with √9x^2 +, is it not suppose to be √9x+
do u have limit at infinity with table of values??
can i know how to solve the question at @14.35 if the limit is at negative infinity
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isn't 9x^2 times 1/x = 9x, not 9? @ 15:54
thats what im confused at aswell
Cuz that’s inside a square root, 1/x becomes 1/x^2
everything that is placed from outside the square root to the inside got squared
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hey at 12:56 why did the square root of 1/x^6 (9x^6 - x^2) turn into square root (9 + ....) which is addition now? where did the addition sign come from??
Same question right here too😂😅
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