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I am currently learning about infinite limits in Calculus, basically determining the limit of a function as x approaches infinity. However, I am struggling to understand the method being used to find it.

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Let’s take the function above. The method above seems to be to ignore all the terms with a lower power except the terms with the highest power. Then, because both of the highest terms are x^5, we cross them out and then divide their coefficients, to get a limit of 2/3.

I guess you can do this because at infinity, the value of x^5 would be so big that it would dominate all the other values However, I still have a few problems with this method:

  1. But on that logic, why should we care that the coefficient of the numerator of is 4 and the denominator’s coefficient is 6? The value of x^5 is so big that it would dominate both of them anyway? At this rate, because it would dominate everything, both the numerator and denominator of every function approaching infinity should be infinity over infinity! So wouldn’t the limit of all infinite functions be 1?

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  1. This is another example that confuses me. Apparently, when x is infinity, you can ignore the 10, because infinity would dominate the whole function, and therefore the limit would be 0. But this doesn’t make sense to me! Even at infinity, the difference between the 2 would be 10, not 0. No matter how large a number you sub in, the difference between the 2 functions will be 10, and therefore, how can the functions approach 0 as x approaches infinity?

  2. I also I get that if you zoom out in the function above, it would truly seem like that function is approaching 0. But then that wouldn’t be the limit of the function would it? That would be zooming out! Once we zoom back in, we will be able see that the function is sticking at 10, not getting closer to 0! So how can we say that the limit of the function at infinity is 10?

Can someone explain the above to me? Can you also not make the explanation too rigorous? I’m just learning Khan Academy Calculus, and still haven’t touched on things like epsilon delta proofs yet. Thank you!

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  • $\begingroup$ $\infty -\infty $ is sort of undetermined $\endgroup$ – Entrepreneur Aug 28 '18 at 6:37
  • $\begingroup$ "the function is sticking at $10$": how so ? $\sqrt{100000100}-\sqrt{100000000}=0.00499999875000\cdots$. $\endgroup$ – Yves Daoust Aug 28 '18 at 7:05
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    $\begingroup$ These examples use shortcuts - they are not meant to be rigorous. Try some numerical examples (plug in increasing values for x) to convince yourselves that these shortcuts produce the correct results in these examples. To understand why the shortcuts work, you need to understand epsilon delta proofs, which are the rigorous way to determine limits. $\endgroup$ – gandalf61 Aug 28 '18 at 8:18
  • $\begingroup$ A simplistic explanation why the lower powers don't matter but the leading numerical coefficients do: as $x$ gets very large, $x^6$ becomes negligibly small compared to $x^7,$ but no matter how large $x$ gets, $9$ will always be three times as big as $3.$ $\endgroup$ – David K Aug 28 '18 at 13:39
  • $\begingroup$ Also concerning your question of zooming out, as you noted a limit is not zooming out. It's rather going further right on the x axis : you look at what's going on near infinity. If you do the calculation as given in the chosen answer, you'll see that indeed, the function goes to $0$. $\endgroup$ – Mariuslp Aug 28 '18 at 14:50
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The logics for the ratio of polynomials:

From a ratio with equal degrees such as

$$\lim_{x\to\infty}\frac{3x^2+2x+1}{2x^2+7x-4}$$

we can rewrite

$$\lim_{x\to\infty}\frac{3+\dfrac2x+\dfrac1{x^2}}{2+\dfrac7x-\dfrac4{x^2}}.$$

In the latter expression, it is clear that the terms with a denominator will vanish and all that remains is $\dfrac32$.

If the degrees differ, the highest degree "wins".

$$\lim_{x\to\infty}\frac{3x^4+2x^2+1}{2x^2+7x-4}=\lim_{x\to\infty}x^2\frac{3+\dfrac2{x^2}+\dfrac1{x^4}}{2+\dfrac7x-\dfrac4{x^2}}=\lim_{x\to\infty}x^2\frac32.$$

$$\lim_{x\to\infty}\frac{3x^2+2x+1}{2x^3+7x-4}=\lim_{x\to\infty}\frac1x\frac{3+\dfrac2x+\dfrac1{x^2}}{2+\dfrac7{x^2}-\dfrac4{x^3}}=\lim_{x\to\infty}\frac1x\frac32.$$


Note that even if the numerator and denominator both tend to infinity, their ratio needn't be $1$, because as they grow, their ratio can be very different from $1$. This is the very spirit of limits: you observe the behavior at some point (which can be infinity) by extrapolating from the behavior at nearby points.


The logics for differences:

When you subtract two quantities, you must be very careful because there can be cancellation: if the two quantities are close to each other, the difference can become significant.

For instance, in

$$\lim_{x\to\infty}(\sqrt{x+10}-\sqrt x)$$ you may not simplify the $10$ under the square root saying that it is negligible in front of $x$, because the second $x$ will counteract the first. But as we cannot say $\sqrt{x+10}=\sqrt x+\sqrt{10}$ and simplify, the computation needs to be smarter.

The classical way is to write

$$\sqrt{x+10}-\sqrt x=\frac{x+10-x}{\sqrt{x+10}+\sqrt{x}}$$ and now as there is no cancellation in the denominator, this can be replaced by

$$\frac{10}{2\sqrt x}.$$

Another way is to pull $\sqrt x$ out

$$\sqrt{x+10}-\sqrt x=\sqrt x\left(\sqrt{1+\frac{10}x}-1\right),$$ and as $\dfrac{10}x$ is smaller and smaller, you can linearize:

$$\sqrt{1+\epsilon}\approx 1+\frac\epsilon2,$$ with an approximation that is better and better for smaller and smaller $\epsilon$.

Now

$$\sqrt x\left(\sqrt{1+\frac{10}x}-1\right)\approx\sqrt x\left(1+\frac{10}{2x}-1\right)=\frac{10}{2\sqrt x}.$$


The figure illustrates linearization of $\sqrt{1+\epsilon}$ near $0$.

enter image description here

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  • $\begingroup$ Note that the example in the pictures has $\sqrt{x + 100}$. The OP was making the mistake ot thinking $\sqrt{x + 100} = \sqrt x + \sqrt{100} = \sqrt x + 10$. @EthanChan - square roots do not break over addition like that. $\endgroup$ – Paul Sinclair Aug 28 '18 at 16:39
  • $\begingroup$ @PaulSinclair: I said it. $\endgroup$ – Yves Daoust Aug 28 '18 at 18:03
  • $\begingroup$ Not quite - you talk about $\sqrt{x + 10}$, but the OP discusses $\sqrt{x + 100}$. Of course, your analysis is correct other than this small change. But I wanted to point out the difference to Ethan, to make sure the change in the problem didn't hide this particular correction from him. $\endgroup$ – Paul Sinclair Aug 28 '18 at 19:55
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With this kind of thing, when you're learning, get a calculator and put some numbers in. That will help convince you of what's going on (and sometimes surprise you -- some convergences are very, very slow: to get $\sum 1/n$ to reach $10$ you need nearly $30,000$ terms... but it still goes to $\infty$).

Right, specifics: as $x$ gets large, $x^5$ becomes significantly bigger than $x^4, x^3, x^2$ and $x$ -- put some numbers in (say 10, 25, 50, and 100) to see this. Then note that every time you increase the value of $x$ the terms with the highest power get multiplied by that larger number more often, so they must grow faster. Because of that, we say that the term $x^n$ with the largest $n$ dominates the expression, and we can ignore the rest. To be mathematically precise we can multiply through by $x^{-n}$ which will give us a constant term (the co-efficient of $x^n$) and then a collection of terms of the form $(1/x)^n$. Since $x\rightarrow \infty$, $1/x \rightarrow 0$ -- and this is why we don't ignore the co-efficient of $x^n$ in the limit.

When we establish the dominant term in the numerator and the denominator we're left with a much simpler quotient and we can 'cancel' the $x$'s. Actually, we're multiplying top and bottom by $x^{-n}/x^{-n} = 1$ where $n$ is the larger of the two powers. If numerator and denominator have the name power $n$ then we're left with a fraction, and that's the limit of the sequence. Otherwise we have some power of $x$ in either the numerator (so we go to $\infty$) or the denominator (so we go to $0$).

At this point, go back through your examples above and try this. You should find it starts to make more sense.

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As you stated:

I guess you can do this because at infinity, the value of x^5 would be so big that it would dominate all the other values.

This is absolutely right. That's exactly why it works. However, it's crucial to be cautious about this reasoning since there are some things to consider. Particularly you should always consider how each function you are analyzing changes as $x \to \infty$ or $x \to -\infty$.

For the case of polynomials you just need to check the leading term in each polynomial because it can be shown (I'll show you how) that the leading term dominates (as you said).

I'll go to each of the problems you mention one by one:

  1. Say you want $$\lim_{x \to \infty}{3x^4 \over 7x^4}$$ Then, for any value $x \neq 0$ (including huge values as $x \to \infty$) we have that ${3x^4 \over 7x^4} = {3 \over 7}$. Which means that $$\lim_{x \to \infty}{3x^4 \over 7x^4} = \lim_{x \to \infty}{3 \over 7} = {3 \over 7}$$ A simpler way to see it is that you can think of a constant like $3$ as a constant function $f(x) = 3$ for all $x>0$. Then $f$ can also be expressed as $f(x) = {3x \over x}$ (since $f$ is defined just for positive values). Then, $\lim_{x \to \infty} f(x)$ is clearly equal to 3. Say you now want $$\lim_{x \to \infty}{3x^4 - 2x + 3 \over 7x^4 + 8x^3 + 5x^2 + 6}$$ Then, from what we have seen above, all our steps below are now justified except the first one: $$\lim_{x \to \infty}{3x^4 - 2x + 3 \over 7x^4 + 8x^3 + 5x^2 + 6} = \lim_{x \to \infty}{3x^4 \over 7x^4} = \lim_{x \to \infty}{3 \over 7} = {3 \over 7}$$ If you divide both numerator and denominator by the leading term ($x^4$) it will be clear why the first step works (so the method of just taking the leading terms works): $$\eqalign{ \lim_{x \to \infty}{3x^4 - 2x + 3 \over 7x^4 + 8x^3 + 5x^2 + 6} &= \lim_{x \to \infty}{{3x^4 \over x^4} - {2x \over x^4} + {3 \over x^4} \over {7x^4 \over x^4} + {8x^3 \over x^4} + {5x^2 \over x^4} + {6 \over x^4}} \\ &= \lim_{x \to \infty}{3 - {2 \over x^3} + {3 \over x^4} \over 7 + {8 \over x} + {5 \over x^2} + {6 \over x^4}} \\ &= {3 \over 7} }$$ since as $x \to \infty$ all terms except $3$ and $7$ go to $0$.
  2. A first approach to this problem can be to just plug in increasing values for $\sqrt {100 + x} - \sqrt x$. The case for a difference of $10$ can be found when $x = 0$ but what if you plug in $x = 10000$? Or $x = 1000000000$? You will find that this difference will decrease as $x \to \infty$ and it is not always 10 as you thought. This problem is the reason why I suggested to always consider how your functions behave before ignoring terms and those sort of things. The 10 can be ignored because as you can see fromthis image it's not the same to increase the value of $x$ from 0 to 100 than increasing it from 1000000 to 1000100 because of the way $\sqrt x$ works (it is concave down). So this is why if you set $f(x) = \sqrt x$ then $lim_{x \to \infty} f(100 + x) - f(x) = 0$. If you do the same with $f(x) = x$ and with $f(x) = x^2$ you will see a different $\lim$ of $f(100 + x) - f(x)$.
  3. I guess this problem is solved with the answer to 2.
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    $\begingroup$ In point 1 I believe there's a typo, an exponent 3 that should be a 4. I can't edit it due to the "minimum 6 characters" requirement. $\endgroup$ – Todd Sewell Aug 28 '18 at 8:31
  • $\begingroup$ Fixed! Thanks you! $\endgroup$ – salvarico Aug 28 '18 at 15:32
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For 2 and 3, it seems like you are making an algebra mistake.

$$(\sqrt x+10)^2=x+20\sqrt x+100\ne x+100$$

For example, if $x=10000, \sqrt x=100$. $\sqrt{10100}<101$. Indeed, your problem shows $\sqrt{100+x}-\sqrt x=\dfrac{100}{\sqrt{100+x}+\sqrt x}$. As x increases, the denominator increases without bound while the numerator stays the same.

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$1.$

Yes, you’re right; the term with the highest power is so dominant we can sort of ‘ignore’ the other terms. Although, I think the best way to show this is to divide each term by the term of highest degree. In your first example, we would divide each term by $x^7$, and then each term apart from the $9x^7$ and the $3x^7$ would have some degree of $x$ in its denominator, thus would tend to $0$ as $x \rightarrow \infty$.

$2.$

I don’t know what you’re referring to here. There is no $10$ present in the second picture!

$3.$

I don’t know where you’re getting this $10$ from?

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The problem with the thought process is you want the flat answer at infinity, when a mathmatician is looking at infinity we have to see that it is a concept and not attainable so the focus is not what it is but rather what it tends to.

  1. The answer to the first problem is that we only care about a part that matters most. Sice the terms cancel out the thing it tends to most would be the coefficient. This is mainly just a definition we decided on after having graphed many examples and seen the pattern and it holding through. However this really only works for numbers we imagine because at infinity it will get pulled on by the others. So in this case we only care about the largest powers.

  2. The second is harder to see because if you pull them term by term it seems it is ten but the fact that the 100 and x are being subtracted under the square root together brings in more complexity. In his proof of the answer we see that the original doesn't give the whole picture and can be seen to go to 0.

  3. We chose to take a zoomed out veiw of the function because that is what we're looking at. If we zoom in on any one part it seems the answer is something else. These are standards that mathmaticians have set up to keep it regulated. I like to think of this veiw as not zooming in on a section or zooming out but looking at what it tends to. Imagine following the function as it is being graphed and with sufficient length we can accurately predict what the answer will be towards numbers around it. This is why we look towards what it tends to not actually infinity. Infinity is just what we like to refer to that as.

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