Epic Item 53
by EpicSkills32, Apr 18, 2014, 6:33 AM
Here's a cool calculus trick I learned just now.
I'll show it by using it on a problem.
![\[ \lim_{k \to \infty} \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} \]](http://latex.artofproblemsolving.com/9/a/1/9a10caadbdf72dfed438d1094a34eada5a113156.png)
First Look: We take a look to see if there are any possible shortcuts or easy ways to do this. Unfortunately, we don't see any really high powers or anything to cancel. The ln (sounds like "lawn" -> briantix) looks nasty and the square root doesn't make anything better.
In short: ugh
Deeper Analysis: We check out the natural log function. What the function is evaluating is obviously approaching infinity, so we're taking the natural log of an extremely large number. In the denominator, we see a square root function. hm.... How does the square root function compare with the natural log function? I don't know either, so we won't be able to solve this problem by looking at higher values on the top or bottom.
hidden because of length
Here the cool trick comes in. In all of our analysis of the problem, we didn't talk at all about continuity or domain or whatever of the function. Right now, we notice that the function is strictly positive for values of k greater than 0. Why? When k is positive, the numerator is the natural log of something positive, and the square root of a positive value is positive. (Or the preferred value/one we're using)
Thus we can write this:
![\[ \lim_{k \to \infty} \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}}\geq 0 \]](//latex.artofproblemsolving.com/9/f/2/9f29c08030cf8e83f73e96083621eae6fdd6b35a.png)
We looked at the function for when k is positive; can we do anything like that for other restrictions on k? Since the problem is asking for a value of k at (or approaching) infinity, almost anything would work.
hm...
Here's another sort of trick that will boost the trick we already started. Noticing the inequality (we're ignoring the limit right now), we then right another inequality.
WHAT???? Where did that come from?
Explanation: 12+3+1=16. wait what. In reality, the exact number doesn't really matter as long as it's bigger than 16. The important thing is to notice that the inequality is true for values of k greater than 1. When k is exactly 1,
is equal to 
and 
. When k is greater than 1, is greater than and . (Same for k greater than anything up to infinity actually, but that doesn't matter, as you'll see later) Now remember: You cannot just do this for any problem anywhere. We were able to do this because: 1- We're taking the limit of something approaching infinity, and in the long run, only the highest powers matter. 2- This whole expression is greater than or equal to the original expression. Changing to something still positive still makes the inequality true.
For other problems, I would not recommend changing values like that. In this problem, we're not looking for k. In fact, k is changing (increasing indefinitely btw), so when we want a slightly different form or value, something like
is not much different from 
.
Ok anyway: we now have:
![\[ \dfrac{\ln\left(16k^4\right)}{\sqrt{k}} \geq \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} \,\, \texttt{for k>1} \]](//latex.artofproblemsolving.com/e/e/9/ee9384d9790e0605c49a5812fd21d5d6b5b75a05.png)
By simple logarithm rules, we can separate the fraction (and pull the 4 down), and we rewrite the square root as a fractional exponent:
![\[\dfrac{\ln(16)}{k^{\dfrac{1}{2}}} + \dfrac{4\ln(k)}{k^{\dfrac{1}{2}}} \,\,\, \texttt{for k>1} \]](//latex.artofproblemsolving.com/d/c/0/dc0f0d0bdf2035876e64fe1da10d1ca66f99b13a.png)
We now do some analysis. Right off the bat we should notice that the problem is much nicer now. Actually, it's not really a problem. We're simply looking at an expression. But remember: keep in mind the goal What are we looking for again? The original problem was asking for the limit as k got really huge. Right now we have an expression that isn't the exact same, but close. However, remember what we did with the inequality? We're looking at a similar expression in relation to zero. We know that this expression is bigger than zero.
Well the problem is asking for a limit; why not think about that now?
Since the expression we have is for k>1, and infinity is definitely bigger than 1, why not look at this expression around infinity? (we're leaping from 1 to infinity haha)
The limit of a sum is the sum of the limits. What are the limits of the two fractions?
As k gets really big for
, the fraction gets closer and closer to zero. (If this isn't obvious to you, it's because
So that part is essentially zero.
What about when k gets huge for
?
Another important fact: ln grows more slowly than any positive power In other words, x raised to any positive power grows faster (eventually at least) than the natural log of x.
From what we just said^^, the
will eventually, as k grows, be smaller than 
.
So......the limit of all this as k increases indefinitely, is zero.
The sum of these two limits is the same as the limit of the two combined. Now let's think about this. We know two main things: An expression larger than the original has the limit of zero.
![\[ \dfrac{\ln(16)}{k^{\dfrac{1}{2}}} \]](//latex.artofproblemsolving.com/f/2/1/f2181b4d74986f26776bfa9abc3e42cdc1894f6b.png)
approaches zero as k gets infinitely large.
Or:
![\[ \dfrac{\ln(16)}{k^{\dfrac{1}{2}}} \geq \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} \]](//latex.artofproblemsolving.com/d/c/d/dcd4bd9ca9a0811c4a48102af35d530d011a3cf6.png)
![\[ \lim_{k\to\infty}\dfrac{\ln(16)}{k^{\dfrac{1}{2}}} = 0 \]](//latex.artofproblemsolving.com/a/0/a/a0a0e4e0a3363be3fb34dd188f2726bb956c61b4.png)
We also know that the original value is strictly positive for positive values of k, or:
![\[ \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} \geq 0\,\,\, \texttt{for k>0} \]](//latex.artofproblemsolving.com/9/f/d/9fd35f909609b8549f8281bfae692611225264ab.png)
hm.....
How can we turn the second part into something to do with limits?
Well the second expression is the original problem right? And we know that it is always positive when k is positive. This means that it can be as close to zero as we want, as long as it is above zero. hm....this sounds like limits...... It IS!
Because we determined that the original expression is always above zero for positive inputs, we can say that the expression is restricted or "bounded" by zero. This works in this problem because k is approaching infinity, a positive value.
Now what else did we figure out about this expression? When k is greater than 1, (<- remember that has to be true!), the limit as k approaches infinity for this expression is zero.
hm...well infinity is not only positive.....it's greater than 1!!!!!!!! shocker right there.
Finale: The original expression is definitely positive. This does not mean it is a large value; the output could be infinitely close to zero but positive. Thus there is a limit for the expression that equals zero. (Something that makes the function super close to zero, but not quite.)
The new expression, which is greater than the original, has a limit of zero.
From this we write a big expression with everything:
![\[ \dfrac{\ln\left(16k^4\right)}{\sqrt{k}} \geq \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} \geq 0 \]](//latex.artofproblemsolving.com/6/0/6/606f5c23eb2e1fc15dca88e557922d431a3ce025.png)
Now it should be pretty clear how to finish. We have:
Something approaching zero is greater than or equal to a limit we want to find is greater than or equal to zero.
There's a convenient thing in calculus called the squeeze/squeezing/pinching theorem. http://en.wikipedia.org/wiki/Squeeze_theorem
Look at our expression, we see that the limit of our original expression is being "squeezed" between two "limits" or boundaries. The limit greater than it is
, while we know the expression is strictly positive for positive inputs, like infinity.
The expression
is less than 0, but also greater than or equal to 0. (*Technically, but remember limits don't define them as being there, just being close*)
Thus by the squeeze theorem:
![\[ \lim_{k \to \infty} \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} = 0 \]](//latex.artofproblemsolving.com/b/e/7/be7ef406f4cab97b0f80c4a454ca88abe6af7a16.png)
in case you forgot the
Thus we can write this:
![\[ \lim_{k \to \infty} \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}}\geq 0 \]](http://latex.artofproblemsolving.com/9/f/2/9f29c08030cf8e83f73e96083621eae6fdd6b35a.png)
We looked at the function for when k is positive; can we do anything like that for other restrictions on k? Since the problem is asking for a value of k at (or approaching) infinity, almost anything would work.
hm...
Here's another sort of trick that will boost the trick we already started. Noticing the inequality (we're ignoring the limit right now), we then right another inequality.
WHAT???? Where did that come from?
Explanation: 12+3+1=16. wait what. In reality, the exact number doesn't really matter as long as it's bigger than 16. The important thing is to notice that the inequality is true for values of k greater than 1. When k is exactly 1,
is equal to 
and 
. When k is greater than 1, is greater than and . (Same for k greater than anything up to infinity actually, but that doesn't matter, as you'll see later) Now remember: You cannot just do this for any problem anywhere. We were able to do this because: 1- We're taking the limit of something approaching infinity, and in the long run, only the highest powers matter. 2- This whole expression is greater than or equal to the original expression. Changing to something still positive still makes the inequality true. For other problems, I would not recommend changing values like that. In this problem, we're not looking for k. In fact, k is changing (increasing indefinitely btw), so when we want a slightly different form or value, something like
is not much different from 
.Ok anyway: we now have:
![\[ \dfrac{\ln\left(16k^4\right)}{\sqrt{k}} \geq \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} \,\, \texttt{for k>1} \]](http://latex.artofproblemsolving.com/e/e/9/ee9384d9790e0605c49a5812fd21d5d6b5b75a05.png)
By simple logarithm rules, we can separate the fraction (and pull the 4 down), and we rewrite the square root as a fractional exponent:
![\[\dfrac{\ln(16)}{k^{\dfrac{1}{2}}} + \dfrac{4\ln(k)}{k^{\dfrac{1}{2}}} \,\,\, \texttt{for k>1} \]](http://latex.artofproblemsolving.com/d/c/0/dc0f0d0bdf2035876e64fe1da10d1ca66f99b13a.png)
We now do some analysis. Right off the bat we should notice that the problem is much nicer now. Actually, it's not really a problem. We're simply looking at an expression. But remember: keep in mind the goal What are we looking for again? The original problem was asking for the limit as k got really huge. Right now we have an expression that isn't the exact same, but close. However, remember what we did with the inequality? We're looking at a similar expression in relation to zero. We know that this expression is bigger than zero.
Well the problem is asking for a limit; why not think about that now?
Since the expression we have is for k>1, and infinity is definitely bigger than 1, why not look at this expression around infinity? (we're leaping from 1 to infinity haha)
The limit of a sum is the sum of the limits. What are the limits of the two fractions?
As k gets really big for
, the fraction gets closer and closer to zero. (If this isn't obvious to you, it's becausethere's a constant in the numerator, while the bottom is really big. In fact, it's technically infinity. (Infinity to the half power is infinity basically) A small value, natural log of 16 divided by an infinitely large number approaches zero. (Remember: It doesn't necessarily equal zero; it just comes as close as we want it to(and infinitely closer after XP))
So that part is essentially zero.
What about when k gets huge for
?Another important fact: ln grows more slowly than any positive power In other words, x raised to any positive power grows faster (eventually at least) than the natural log of x.
From what we just said^^, the
will eventually, as k grows, be smaller than 
.So......the limit of all this as k increases indefinitely, is zero.
The sum of these two limits is the same as the limit of the two combined. Now let's think about this. We know two main things: An expression larger than the original has the limit of zero.
![\[ \dfrac{\ln(16)}{k^{\dfrac{1}{2}}} \]](http://latex.artofproblemsolving.com/f/2/1/f2181b4d74986f26776bfa9abc3e42cdc1894f6b.png)
approaches zero as k gets infinitely large.Or:
![\[ \dfrac{\ln(16)}{k^{\dfrac{1}{2}}} \geq \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} \]](http://latex.artofproblemsolving.com/d/c/d/dcd4bd9ca9a0811c4a48102af35d530d011a3cf6.png)
![\[ \lim_{k\to\infty}\dfrac{\ln(16)}{k^{\dfrac{1}{2}}} = 0 \]](http://latex.artofproblemsolving.com/a/0/a/a0a0e4e0a3363be3fb34dd188f2726bb956c61b4.png)
We also know that the original value is strictly positive for positive values of k, or:
![\[ \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} \geq 0\,\,\, \texttt{for k>0} \]](http://latex.artofproblemsolving.com/9/f/d/9fd35f909609b8549f8281bfae692611225264ab.png)
hm.....
How can we turn the second part into something to do with limits?
Well the second expression is the original problem right? And we know that it is always positive when k is positive. This means that it can be as close to zero as we want, as long as it is above zero. hm....this sounds like limits...... It IS!
Because we determined that the original expression is always above zero for positive inputs, we can say that the expression is restricted or "bounded" by zero. This works in this problem because k is approaching infinity, a positive value.
Now what else did we figure out about this expression? When k is greater than 1, (<- remember that has to be true!), the limit as k approaches infinity for this expression is zero.
hm...well infinity is not only positive.....it's greater than 1!!!!!!!! shocker right there.
Finale: The original expression is definitely positive. This does not mean it is a large value; the output could be infinitely close to zero but positive. Thus there is a limit for the expression that equals zero. (Something that makes the function super close to zero, but not quite.)
The new expression, which is greater than the original, has a limit of zero.
From this we write a big expression with everything:
![\[ \dfrac{\ln\left(16k^4\right)}{\sqrt{k}} \geq \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} \geq 0 \]](http://latex.artofproblemsolving.com/6/0/6/606f5c23eb2e1fc15dca88e557922d431a3ce025.png)
Now it should be pretty clear how to finish. We have:
Something approaching zero is greater than or equal to a limit we want to find is greater than or equal to zero.
There's a convenient thing in calculus called the squeeze/squeezing/pinching theorem. http://en.wikipedia.org/wiki/Squeeze_theorem
Look at our expression, we see that the limit of our original expression is being "squeezed" between two "limits" or boundaries. The limit greater than it is
, while we know the expression is strictly positive for positive inputs, like infinity.The expression
is less than 0, but also greater than or equal to 0. (*Technically, but remember limits don't define them as being there, just being close*)Thus by the squeeze theorem:
![\[ \lim_{k \to \infty} \dfrac{\ln\left(12k^4+3k+1\right)}{\sqrt{k}} = 0 \]](http://latex.artofproblemsolving.com/b/e/7/be7ef406f4cab97b0f80c4a454ca88abe6af7a16.png)
in case you forgot the
We basically wrote the expression in an inequality and found two things to stick it between. Evaluating limits of those two things, we found the correct limit using the squeeze theorem.
This post has been edited 3 times. Last edited by EpicSkills32, Apr 18, 2014, 6:04 PM
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