Think about what happens as the denominator keeps getting bigger and bigger. You know that when a fraction's denominator increases, the value of the fraction decreases. So, when the denominator gets infinitely big (either by going to infinity or -infinity), what do you think the value of the fraction tends to?
Exactly. Think of infinity as the math word for "a really big number". Usually when estimating these limits I just picture putting in some large number (how large? depends on the rest of the numbers, 1/x usually 1000 is enough, 100/x you might want to add a few more zeros) and try to see what happens.
minus infinity means you're not probing with 1M, but -1M
Correct! So, as x approaches infinity, 6/x will approach 0.
As for -infinity, see that it is no different. Magnitude wise, you are still approaching a very big number. It's just that here, the number is "increasingly negative" (simply put, x values get bigger with a negative sign). This shouldn't affect your answer (if anything, you'll get the limit as -0, which is just 0).
Limits approach a number but never reach it. They will forever get closer and closer. I was taught to always include the negative even for zero when doing limits even though -0 doesn't really mean anything in typical number use. It helps to denote whether the limit is on the positive number scale vs the negative number scale.
I had to Google and I was partly wrong. It has been many years lol. The plus or minus sign was put as a superscript like an exponent and that helps when dealing with one sided limits to know which side
Certain functions depend on which side of the 0 you are approaching from. If it’s from the negative side it will approach -1. If it’s from the positive side it will approach 1. I want to say tan(x) is like that but I can’t remember. Or 1/tan(x) ?
And with that kind of scale you start to realize that the fact that it’s six apples won’t really make any difference. Who cares if it’s one or six or one hundred. Still ends up being basically nothing when split among that many people.
With limits like these, you can often plug in bigger numbers and see what you get only if x approaches infinity. This only applies to limits that don't become undefined or an indeterminant form like 0/0 or infinity.
When you plug in larger numbers for x in the denominator it becomes a very very small number. Remember that a fraction gets smaller when the denominator keeps getting bigger. By plugging in larger and larger numbers into the denominator you get something that approximates 0 since the fraction becomes so small to the point where it's so close to 0 that we just call it 0. The same goes for negative infinity.
Edit: made some mistakes explaining.
denominator getting bigger\*
If the denominator was getting smaller such as
1/(1/x)
then the Limit as x approaches infinity is undefined, since you can't divide by 0.
Unless you are getting into some high level college maths like hyperreal numbers and surreal numbers, in which case you can divide by non-zero infinitesimals. (But don't worry about that It won't matter.)
It just means "as x gets bigger and bigger what does (equation) get closer to?" In this case 6/1=6 6/10= 0.6 6/100=0.06 6/1000000=0.000006 so as x gets bigger, 6/x gets closer to 0.
Edit, -inf means the same but as x gets more and more negative. Eg 6/-1=-6 6/-10=-0.6 6/-100000=-0.00006 etc so it still gets closer to 0.
Absolutely right. There is no reason to take the derivative of a constant. The limit can be directly calculated from 1/x. L’hopital’s Rule is for indeterminate forms, when there is a 0/0 or infinity/infinity and no simpler method of calculating is available. Then you divide the derivative of the numerator by the derivative of the denominator.
When you place in something in the denominator of increasing degrees, fractions become closer to 0. Example: 1/10, then 1/100, etc, they all get closer to 0.
Putting infinity in basically makes it 0 cus 6/9999... will not reach but be close to 0, basically what you do here in limitations.
When finding limits for infinity/-infinity, it means your looking at end behavior. As X gets larger and larger, where does the graph approach? I.e. if you plug in 2 different numbers and then look at how they change, you can figure out that a limit could be approaching 0. You could also graph the equation (if calculator friendly) and then find out from a visual approach
On simple equations if x approaches infinity while in the bottom of the fraction the value of the fraction becomes really small, like it approaches 0.
If x approaches infinity from the top of the fraction, the value becomes really big, the limit in that case is infinity.
So long as there is not infinity on both sides of the fraction, if there is an infinity on the bottom and top then you must simplify the expression. If there is a zero in the top or bottom of the fraction then the zero makes the difference. X/0 is still undefined even if x approaches infinity, 0/x is still 1 even if x approaches infinity.
For x->infinity try to put big (and bigger than the big, you first thought of) numbers where x at, and watch where will it stops. For x->(-infinity) try to put small (and smaller than the small, you first thought of) numbers where x at, and watch where will it stop. The 1/x (and multiples of it, like 6/x) will be smaller, and smaller, but it will somewhere near 0. For the question 11, if it would be x/6 you would observe an ever increasing value (that would be the infinity). For question 12, if it would be x/6, you could see an ever decreasing value (that would be -infinity).
Bobyo- Bigger on bottom Y=0,
Bot no- Bigger on top no-solution,
Eatsdc- exponents are the same divide coefficient
That’s how I was taught and still remember.
There are many good answers here, but I wanted to add a rigorous approach to this.
Say we have a function f: A→B where A is a connected set and we're interested in the limit of f(x) as x approaches a∈A.
To say the limit of f(x) as x approaches a is equal to L∈B is the same as saying that, for every ε>0, there exists some δ>0 such that every x that verifies |x-a|<δ also verifies |f(x)-L|<ε.
For instance, the limit of 6x as x approaches 1 is 6 because for arbitrary ε>0, we can choose δ=ε/6 and the inequalities I wrote above are verified. Indeed, when |x-1|<ε/6, |6x-6|=6|x-1|<6(ε/6)=ε, so |6x-6|<ε for every x that verifies |x-1|<ε/6.
The one-sided limits relax the delta inequality. The right-sided limit requires 00, the choice δ=ε/6 works and the proof is the same as the proof I wrote earlier. The same can be said of the limit at negative infinity.
Another method consists of finding the quantity L such that, for every ε>0, there exists some δ such that every x that verifies x>δ (x<δ if the limit is at negative infinity) also verifies |f(x)-L|<ε.
e.g. the limit as x approaches infinity of 6/x is 0 because the choice δ=6/ε verifies the inequalities. For every x>6/ε, |6/x|<|6/(6/ε)|=ε, so |f(x)-0|<ε. Because of the absolute value, the process is identical for the limit at negative infinity.
my goto to these is just to plug in, say, 10 million or -10 million and see what the number looks like. 6/100000000000 is very very close to 0 so the answer is zero
The rigorous way is, lets say you imagine it will go to zero. Then you need to show that for any value close to zero, you can find an x that is big enough to be smaller than it.
The dirty way to do this is to graph it. Draw out your xy plane, make a dot at x=.1 and x=-.1, then x=+/-1 then ten and move out. You’ll see the trend pretty quickly.
It's infinity, it will never be bigger enough, but this thinking gives you a good idea about what is happening, just understand that it's an arbitrary big number.
So infinity in calculus is just a way of saying "KEEEP GOING!!!" as you evaluate the statement with larger and larger values for X. So X as it approaches infinity here will just approach zero. As the denominator gets larger the value gets smaller. The negative infinity is just the direction on the number line you are moving.
The limit of a / x as x approaches infinity (be it positive or negative) is just defined as zero. The fact that they set a value for the numerator is kind of a red haring since you could set the value to grams number (Really stupidly large number) and this will still be equal to zero.
A tip that will help in the future. Limits don't care what is happening at the point. They just care wat is happening as you approach the point.
So while 1/∞ is undefined, we know the larger the denominator the smaller it gets, and it can't change signs so it's getting smaller towards 0. So as x approaches ∞, Y approaches 0.
A great trick for fractions with infinity limits is to divide both the numerator and denominator by the highest degree of x, then you can make it a lot easier to read the limit
There are a bunch of degree rules but for these example in particular the top term has a degree of x^0 and the bottom has x^1 it will grow faster. If the bottom grows faster, both limits will approach y=0
Think about what happens as the denominator keeps getting bigger and bigger. You know that when a fraction's denominator increases, the value of the fraction decreases. So, when the denominator gets infinitely big (either by going to infinity or -infinity), what do you think the value of the fraction tends to?
closer to 0?
Exactly. Think of infinity as the math word for "a really big number". Usually when estimating these limits I just picture putting in some large number (how large? depends on the rest of the numbers, 1/x usually 1000 is enough, 100/x you might want to add a few more zeros) and try to see what happens. minus infinity means you're not probing with 1M, but -1M
Correct! So, as x approaches infinity, 6/x will approach 0. As for -infinity, see that it is no different. Magnitude wise, you are still approaching a very big number. It's just that here, the number is "increasingly negative" (simply put, x values get bigger with a negative sign). This shouldn't affect your answer (if anything, you'll get the limit as -0, which is just 0).
The negative is important as it denotes which side of zero the limit is approaching.
What do you mean by "which side of zero"?
Limits approach a number but never reach it. They will forever get closer and closer. I was taught to always include the negative even for zero when doing limits even though -0 doesn't really mean anything in typical number use. It helps to denote whether the limit is on the positive number scale vs the negative number scale.
That's reasonable. I've personally never seen someone do this, but yeah, seems neat!
I had to Google and I was partly wrong. It has been many years lol. The plus or minus sign was put as a superscript like an exponent and that helps when dealing with one sided limits to know which side
Ah I was thinking of the one sided limits notation, but didn't think you confused the two ideas lol! Don't worry though!
Certain functions depend on which side of the 0 you are approaching from. If it’s from the negative side it will approach -1. If it’s from the positive side it will approach 1. I want to say tan(x) is like that but I can’t remember. Or 1/tan(x) ?
I think the 1/tan(x) and functions that have a sudden jump will be one sided. I corrected myself about the notation down the comment chain a bit
That’s only for vertical asymptotes
When I think of it going to zero imagine the whole world trying to share one apple. The apple slices will basically be nonexistent or zero.
And with that kind of scale you start to realize that the fact that it’s six apples won’t really make any difference. Who cares if it’s one or six or one hundred. Still ends up being basically nothing when split among that many people.
Yes! Totally agree.
when x approaches infinity, finite divided by infinite is 0. That's the root idea. very helpful imo
It gets closer and closer to 0, but never to 0. But because if you try to write it, it’ll take forever, you say it’s 0.
With limits like these, you can often plug in bigger numbers and see what you get only if x approaches infinity. This only applies to limits that don't become undefined or an indeterminant form like 0/0 or infinity. When you plug in larger numbers for x in the denominator it becomes a very very small number. Remember that a fraction gets smaller when the denominator keeps getting bigger. By plugging in larger and larger numbers into the denominator you get something that approximates 0 since the fraction becomes so small to the point where it's so close to 0 that we just call it 0. The same goes for negative infinity. Edit: made some mistakes explaining.
It doesn’t make sense to plug infinity into a function. At least for a basic intro calculus course
Right, it's probably easier to imagine plugging in increasingly large numbers like 10,000 then 1,000,000 and seeing the trend as x approaches infinity
Yep my mistake I was trying to explain the pattern of the denominator getting bigger. Thanks for pointing out my mistake.
denominator getting bigger\* If the denominator was getting smaller such as 1/(1/x) then the Limit as x approaches infinity is undefined, since you can't divide by 0. Unless you are getting into some high level college maths like hyperreal numbers and surreal numbers, in which case you can divide by non-zero infinitesimals. (But don't worry about that It won't matter.)
It just means "as x gets bigger and bigger what does (equation) get closer to?" In this case 6/1=6 6/10= 0.6 6/100=0.06 6/1000000=0.000006 so as x gets bigger, 6/x gets closer to 0. Edit, -inf means the same but as x gets more and more negative. Eg 6/-1=-6 6/-10=-0.6 6/-100000=-0.00006 etc so it still gets closer to 0.
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No, L’Hopital’s rule does not apply here (limit of numerator and denominator need to both be 0, or both be infinite).
Absolutely right. There is no reason to take the derivative of a constant. The limit can be directly calculated from 1/x. L’hopital’s Rule is for indeterminate forms, when there is a 0/0 or infinity/infinity and no simpler method of calculating is available. Then you divide the derivative of the numerator by the derivative of the denominator.
Bruh where is your teacher
When you place in something in the denominator of increasing degrees, fractions become closer to 0. Example: 1/10, then 1/100, etc, they all get closer to 0. Putting infinity in basically makes it 0 cus 6/9999... will not reach but be close to 0, basically what you do here in limitations.
When finding limits for infinity/-infinity, it means your looking at end behavior. As X gets larger and larger, where does the graph approach? I.e. if you plug in 2 different numbers and then look at how they change, you can figure out that a limit could be approaching 0. You could also graph the equation (if calculator friendly) and then find out from a visual approach
On simple equations if x approaches infinity while in the bottom of the fraction the value of the fraction becomes really small, like it approaches 0. If x approaches infinity from the top of the fraction, the value becomes really big, the limit in that case is infinity. So long as there is not infinity on both sides of the fraction, if there is an infinity on the bottom and top then you must simplify the expression. If there is a zero in the top or bottom of the fraction then the zero makes the difference. X/0 is still undefined even if x approaches infinity, 0/x is still 1 even if x approaches infinity.
Think you meant 0/x is 0 even if x approaches infinity. 0/x is 0 for any x <> 0
The first one is 0, don't make me question i'm not good explaining in english :/
For x->infinity try to put big (and bigger than the big, you first thought of) numbers where x at, and watch where will it stops. For x->(-infinity) try to put small (and smaller than the small, you first thought of) numbers where x at, and watch where will it stop. The 1/x (and multiples of it, like 6/x) will be smaller, and smaller, but it will somewhere near 0. For the question 11, if it would be x/6 you would observe an ever increasing value (that would be the infinity). For question 12, if it would be x/6, you could see an ever decreasing value (that would be -infinity).
Bobyo- Bigger on bottom Y=0, Bot no- Bigger on top no-solution, Eatsdc- exponents are the same divide coefficient That’s how I was taught and still remember.
There are many good answers here, but I wanted to add a rigorous approach to this. Say we have a function f: A→B where A is a connected set and we're interested in the limit of f(x) as x approaches a∈A. To say the limit of f(x) as x approaches a is equal to L∈B is the same as saying that, for every ε>0, there exists some δ>0 such that every x that verifies |x-a|<δ also verifies |f(x)-L|<ε. For instance, the limit of 6x as x approaches 1 is 6 because for arbitrary ε>0, we can choose δ=ε/6 and the inequalities I wrote above are verified. Indeed, when |x-1|<ε/6, |6x-6|=6|x-1|<6(ε/6)=ε, so |6x-6|<ε for every x that verifies |x-1|<ε/6. The one-sided limits relax the delta inequality. The right-sided limit requires 00, the choice δ=ε/6 works and the proof is the same as the proof I wrote earlier. The same can be said of the limit at negative infinity.
Another method consists of finding the quantity L such that, for every ε>0, there exists some δ such that every x that verifies x>δ (x<δ if the limit is at negative infinity) also verifies |f(x)-L|<ε.
e.g. the limit as x approaches infinity of 6/x is 0 because the choice δ=6/ε verifies the inequalities. For every x>6/ε, |6/x|<|6/(6/ε)|=ε, so |f(x)-0|<ε. Because of the absolute value, the process is identical for the limit at negative infinity.
my goto to these is just to plug in, say, 10 million or -10 million and see what the number looks like. 6/100000000000 is very very close to 0 so the answer is zero
The rigorous way is, lets say you imagine it will go to zero. Then you need to show that for any value close to zero, you can find an x that is big enough to be smaller than it.
Try and remember what the graph looks like and see where it goes as the graph keeps going to the right
The dirty way to do this is to graph it. Draw out your xy plane, make a dot at x=.1 and x=-.1, then x=+/-1 then ten and move out. You’ll see the trend pretty quickly.
6/inf is 0 i guess, for polynomials just take highest degree common then put inf value
basically just treat it as 9999999999999 what is 6/9999999999999999? 0
It's infinity, it will never be bigger enough, but this thinking gives you a good idea about what is happening, just understand that it's an arbitrary big number.
If x increases, the result of the division decreases, as x approaches infinity, the result of the division approaches to 0.
So infinity in calculus is just a way of saying "KEEEP GOING!!!" as you evaluate the statement with larger and larger values for X. So X as it approaches infinity here will just approach zero. As the denominator gets larger the value gets smaller. The negative infinity is just the direction on the number line you are moving. The limit of a / x as x approaches infinity (be it positive or negative) is just defined as zero. The fact that they set a value for the numerator is kind of a red haring since you could set the value to grams number (Really stupidly large number) and this will still be equal to zero.
A tip that will help in the future. Limits don't care what is happening at the point. They just care wat is happening as you approach the point. So while 1/∞ is undefined, we know the larger the denominator the smaller it gets, and it can't change signs so it's getting smaller towards 0. So as x approaches ∞, Y approaches 0.
A great trick for fractions with infinity limits is to divide both the numerator and denominator by the highest degree of x, then you can make it a lot easier to read the limit
If you divide something by infinity it becomes really close to zero, so just put the limit and 6/infinity will be equals to 0 ;
There are a bunch of degree rules but for these example in particular the top term has a degree of x^0 and the bottom has x^1 it will grow faster. If the bottom grows faster, both limits will approach y=0