How to Find Limits
Yilong Yang
October 22, 2014
Contents
1 The General Guideline
2 Put
2.1
2.2
2.3
2.4
1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
2
2
3
4
4
3 Simplification Tricks
3.1 Polynomial over Polynomial and the division trick . .
3.2 Advanced division trick [Optional] . . . . . . . . . . .
3.3 Trig functions . . . . . . . . . . . . . . . . . . . . . . .
3.4 Conjugate trick for square roots . . . . . . . . . . . . .
3.5 Advanced Conjugate tricks for higher roots [Optional]
3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
5
5
5
7
9
9
10
4 Substitution [Optional]
4.1 Linear subs [Optional] . . . . . . . . . . .
4.2 Sub to change the limit [Optional] . . . .
4.3 Sub to change ugly expression [Optional] .
4.4 Universal trig formula [Optional] . . . . .
4.5 Exercises . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
11
11
11
12
13
13
5 Sandwich Theorem
5.1 How to find a good sandwich .
5.2 Advanced sandwich [Optional]
5.3 Sandwich to infinity [Optional]
5.4 Exercises . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
13
13
15
15
16
1
Fractions Together and Factorization
Why put fractions together . . . . . . . .
Formula for factorization . . . . . . . . . .
Advanced factorization [Optional] . . . . .
Exercises . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
The General Guideline
This article serves to illustrate and to summarize the most important tricks used to find limits (without
using L’Hospital Rule). This was originally written for the preparation of the first midterm of MATH 31A.
However, I’m afraid that I’m not very good at designing the difficulty level of problems. Some fo the exercises
in this note might be too hard, and if you can’t solve some of them, just don’t worry about it. I bet the
exam will be easier (or much easier) than those hard problems.
For the tricks listed in this article, I would recommand you to use them in the following order. But you
can of course use your own best judgement about which trick to use and in what order.
1
Procedure 1.1. I recommand using the following procedure when you are asked to find a limit:
1. See if you can plug-in directly (That is, check if the function is continuous at the point of interest.)
2. (Substitution to change limits if necessary)
3. Put fractions together as much as possible, and then factorize if possible.
4. Use appropriate simplifying tricks. (for Polynomials over Polynomials, or for trig functions, or for
square roots, etc.)
5. Use substitution wisely.
6. Try Sandwich Theorem for luck.
In the following sections, I’ll try to explain each steps, the tricks involved, the formula that you should
remember, and provide some exercises or typical problems. When a problem is marked with “(HARD)”,
then it is perfectly Okay if you cannot solve it. When a section is marked with “[Optional]”, then it is
perfectly Okay if you simply skip the section altogether.
As for the answers of these problems, I personally don’t think that they are necessary. As long as you
know how to solve them, it makes no difference to me if you make any calculation mistakes. But if you must
know then feel free to send me an email about which answer you would like to see.
2
2.1
Put Fractions Together and Factorization
Why put fractions together
When you see various fractions flying around (adding together, substracting each other, etc.), you should
put them together using the following formula:
Formula 2.1.
c
ad + bc
a
+ =
b
d
bd
a
c
ad − bc
− =
b
d
bd
Sometimes, there might be hidden “factors” that can be cancelled out, but they only reveal themselves
after you put all the fractions together. For example, we were asked to solve this limit in our homework:
Problem 2.2.
lim
θ→ π
2
1
2
+
2
cos(2θ) + 1 cos (2θ) − 1
Now we should put these fractions together. Let us do this the foolhardy way, using the formula in 2.1.
We see the following.
2
1
2
+
2
cos(2θ) + 1 cos (2θ) − 1
cos2 (2θ) − 1 + 2(cos(2θ) + 1)
=
(cos(2θ) + 1)(cos2 (2θ) − 1)
cos2 (2θ) + 2 cos(2θ) + 1
=
(cos(2θ) + 1)(cos2 (2θ) − 1)
(cos(2θ) + 1)2
=
(cos(2θ) + 1)(cos2 (2θ) − 1)
(cos(2θ) + 1)2
=
(cos(2θ) + 1)(cos(2θ) + 1)(cos(2θ) − 1)
1
=
cos(2θ) − 1
And then since this final expression is continuous at θ =
the answer is − 12 .
2.2
π
2,
we can simply plug-in θ =
π
2,
and see that
Formula for factorization
As you can probably see, the most important thing here is how to factorize an expression. Here are some
really important formula that you should know.
Formula 2.3.
(x + 1)2 = x2 + 2x + 1
(x − 1)2 = x2 − 2x + 1
(x + 1)(x − 1) = x2 − 1
(a + b)2 = a2 + b2 + 2ab
(a − b)2 = a2 + b2 − 2ab
(a + b)(a − b) = a2 − b2
(x + a)(x + b) = x2 + (a + b)x + ab
(x − a)(x − b) = x2 − (a + b)x + ab
Remember here that in the following formula, x, a, b might be changed into any sort of monstrosity. For
example, in our problem 2.2, we see that cos2 (2θ) + 2 cos(2θ) + 1 = (cos(2θ) + 1)2 , and that cos2 (2θ) − 1 =
(cos(2θ) + 1)(cos(2θ) − 1).
It can be super useful if you can factor most degree 2 polynomials by heart. I might or might not write
up something about all the tricks of factoring degree 2 polynomials. Here I shall simply give you a bunch of
problems to work on at the end of this section.
3
2.3
Advanced factorization [Optional]
Finally, I’d like to remark one more use of factorizations. Any substraction is in fact a factorization. To be
more specific, we have the following important formula.
Formula 2.4.
√ √
√
√
a − b = ( a + b)( a − b)
This is a fast way to get rid of the square roots sometimes. For example, you might be asked to solve
the following equation.
Problem 2.5.
lim √
x→3
x−3
x+1−2
Of course, you can use the conjugate tricks as introduced in Section 3.4, or you can also use substitution
as introduced in Section 4.3. But here is another way of showing this.
Since x − 3 is a substraction, therefore
there must be a factorization for it. We would of course want to
√
factorize it into something involving x + 1 (as in the denominator). So we observe the following:
x−3
=(x + 1) − 4
√
√
=( x + 1 − 2)( x + 1 + 2)
With this factorization, we can solve the problem easily.
x−3
lim √
x→3
x+1−2
√
√
( x + 1 − 2)( x + 1 + 2)
√
= lim
x→3
x+1−2
√
= lim ( x + 1 + 2)
x→3
=4
2.4
Exercises
Find all the limits below. Try to use factorization if you can figure out how.
Problem 2.6.
x2 − 5x + 6
x→3 x2 − 4x + 3
lim
Problem 2.7.
x3 − 6x2 + 12x − 8
x→2
x2 − 4x + 4
lim
Problem 2.8.
e2 sin(ln x) − 2esin(ln x) + 1
x→1 e2 sin(ln x) − 8esin(ln x) + 7
lim
Problem 2.9.
lim
x→−2
2
8
+ 2
x+2 x −4
4
Problem 2.10 (HARD).
lim
x→π
2 cot x
1
+
cos x + 1
sin x
Problem 2.11.
lim √
x→−4
Problem 2.12.
lim
x→π
3
x+4
x + 13 − 3
1 − esin x
√
2 − esin x + 3
Simplification Tricks
There are various simplification tricks for various situations. So although this section strive to be as comprehensive as possible, if you find a situation that is not encompassed by this section, please send me an
email about it.
3.1
Polynomial over Polynomial and the division trick
After step one where we put all fractions together, we might be looking at a single fraction where both the
numerator and the denominator are polynomials, and x goes to infinity. For example, we might see this.
Problem 3.1.
2x3 − 3x + 1
+ 2x2 + x − 4
lim
x→∞ 5x3
In this case, we see that x3 outgrows everything else, so we should only care about the coefficients of x3 ,
and read off the answer 25 .
In this case, there is a trick to make this intuition more rigorous. We can divide both the numerator and
the denominator by x3 . Then we are left with the expression:
2 − 3/x2 + 1/x3
5 + 2/x + 1/x2 − 4/x3
Then if we take the limit as x → ∞, we see that everything goes to 0 except for the 2 in the numerator
and the 5 in the denominator (we are implicitly using the limit law here). So the answer is 52 .
3.2
Advanced division trick [Optional]
When doing mathematics, you should always have a simple faith: whenever you have an intuition, then
there must be a corresponding rigorous procedure to prove the intuition. And one of the most fundamental
intuition about doing limits is that, as x goes to ∞, some function will outgrow some other functions. And
the division trick combined with the sandwich theorem, as shown in Section 3.1, is one way to make this
rigorous.
For example, we might see the following problem.
Problem 3.2.
3x + 5 sin x
x→∞ 8x − 6 sin x
lim
The intuition here is that as x goes to ∞, the function f (x) = x must outgrow the function g(x) = sin x
a lot. So, according to our intuition, we are suppose to “ignore” the sin x parts, and only care about the
coefficient of x. Then we can read off the answer as 83 .
5
To make this rigorous, what we do is to first observe the following using the division trick to divide both
numerator and the denominator by x:
3 + 5 sinx x
3x + 5 sin x
=
8x − 6 sin x
8 − 6 sinx x
Then I claim that we indeed have limx→∞ sinx x = 0. Assume that this is true for the moment, then we
can see (by using the limit law) that we can do the following.
3x + 5 sin x
8x − 6 sin x
3 + 5 sinx x
= lim
x→∞ 8 − 6 sin x
x
lim
x→∞
3 + 5 limx→∞
8 − 6 limx→∞
3
3+0
=
=
8−0
8
=
sin x
x
sin x
x
(The Limit Law)
Now, the only thing left to do is to figure out limx→∞
inequality:
0≤
sin x
x .
This is sandwich theorem. Consider the
sin x
1
≤
x
x
Since this inequality is always true, we can take limit as x goes to infinity, and the inequality should be
preserved. So we would have
0 ≤ lim
x→∞
sin x
1
≤ lim
=0
x→∞
x
x
So we indeed have limx→∞ sinx x = 0, and we are done.
The general idea is that, if a function f (x) outgrows a function g(x) by a lot, then we can use some
sandwich theorem with the following sandwich:
0≤
g(x)
≤ some smart choice
f (x)
From there we can deduce that limx→∞ fg(x)
(x) = 0.
Then with the division trick, we will be able to show the limit of the ratio of stuff involving f(x) and g(x).
As another illustration, consider the problem.
Problem 3.3.
3ex − x
x→∞ −4ex + 5x
lim
Since ex will definitely outgrow x, we can simply read off its coefficients and get the answer − 43 . We only
need to show limx→∞ exx = 0. We can use the sandwich theorem in the following way.
First, since we only care about x → ∞, we can assume that x > 2 from now on. (The number 2 here
is an arbitrary choice. I might as weel use 3,4,5,π, or whatever number, as long as the following inequality
holds under such a condition.)
Then the following inequality should hold:
6
ex > x2
Then we see that the following sandwich will be true:
0≤
x
x
≤ 2
ex
x
Now take x → ∞ on everything, and we see that limx→∞
3.3
x
ex
= 0.
Trig functions
The only useful trig functions ever are sin x, cos x and tan x. Out of the three, the more useful ones are sin x
and cos x. And your best friend here should be sin x.
What I’m trying to stress is this: Change everything into sin, and usually wonder will happen.
The reason is the following super important formula that you should remember.
Formula 3.4.
sin x
=1
x→0 x
lim
This formula tells us that sin(blah) and blah are practically the same when this blah goes to 0. For
example, we have the following bunch of formulas, just to name a few.
Formula 3.5.
lim
x→0
sin(2x)
=1
2x
sin(x2 )
=1
x→0
x2
lim
lim
x→−3
sin(x + 3)
=1
x+3
sin(ex )
=1
x→−∞ (ex )
lim
limπ
x→ 2
sin(cos x)
=1
cos x
Now suppose you transformed everything into sin. Then you will see something like the following problem:
Problem 3.6.
sin2 (x) sin(3x)
x→0 x sin(5x) sin(4x)
lim
The trick to tackle this is to replace each sin(blah) by sin(blah)
· blah. Then after using the limit law, this
blah
will simply become 1 · blah, which is blah itself.
Let do this in detail. We have the following derivations.
7
sin2 (x) sin(3x)
x→0 x sin(5x) sin(4x)
lim
1
1
1
= lim [(sin x)(sin x)(sin(3x))
]
x→0
x
sin(5x)
sin(4x)
sin(x)
sin(3x)
1
5x
1
4x
1
sin(x)
· x)(
· x)(
· 3x) (
·
)(
·
)]
= lim [(
x→0
x
x
3x
x sin(5x) 5x sin(4x) 4x
1 1
1
= lim (x · x · (3x) · ·
·
) (This is the limit law)
x→0
x 5x 4x
3
3x3
=
= lim
x→0 20x3
20
Now how exactly should we change all other trig functions into sin x? It is simple to change all trig
functions in term of cos x and sin x. So we only need to remember the following formula to change cos x into
sin x.
Formula 3.7.
1. cos2 x = 1 − sin2 x
2. cos(2x) = 1 − 2 sin2 x
3. cos x = 1 − 2 sin2 ( x2 )
4. cos( π2 − x) = sin x
5. cos( π2 + x) = − sin x
For the sake of completeness, here are some other trig identities you should remember.
Formula 3.8.
1. sin(x + π) = − sin x
2. sin(x − π) = − sin x
3. sin(−x) = − sin x
4. sin(x + π2 ) = cos x
5. sin(x − π2 ) = − cos x
6. cos(x + π) = − cos x
7. cos(x − π) = − cos x
8. cos(−x) = cos x
9. sin(2x) = 2 sin x cos x
10. sin(a + b) = sin a cos b + sin b cos a
11. cos(a + b) = cos a cos b − sin a sin b
12. tan(x + π) = tan(x)
13. tan(x − π) = tan(x)
14. tan(x + π2 ) = − tan1 x
15. tan(x − π2 ) = − tan1 x
8
16. tan(2x) =
2 tan x
1−tan2 x
tan a+tan b
17. tan(a + b) = − 1−tan
a tan b
Finally we have the “universal trig formula”.
Formula 3.9 (Universal trig formula). Let us use the letter t to denote tan( x2 ). Then we have the following:
1. sin x =
2t
1+t2
2. cos x =
1−t2
1+t2
3. tan x =
2t
1−t2
The full power of the universal trig formula lies in the substitution. We shall talk about this in later
sections.
3.4
Conjugate trick for square roots
The conjugate trick for square root is essentially a smart use of the formula (a + b)(a − b) = a2 − b2 . A
typical problem may look like this:
Problem 3.10.
√
x+9−3
2x
√
To get rid of the square root, we can multiply x + 9 + 3 on the numerator and the denominator. We
would have:
lim
x→0
√
x+9−3
x→0
2x
√
√
( x + 9 − 3)( x + 9 + 3)
√
= lim
x→0
2x( x + 9 + 3)
(x + 9) − 9
√
= lim
x→0 2x( x + 9 + 3)
x
√
= lim
x→0 2x( x + 9 + 3)
1
1
= lim √
=
x→0 2( x + 9 + 3)
12
lim
Here we can plug in the value
x = 0 because the last expression
is continuous at x = 0.
√
√
Basically, when
you
see
blah
+
blee,
then
multiply
blah
−
blee
on the numerator and denominator.
√
√
When you see blah − blee, then multiply blah + blee on the numerator and denominator.
3.5
Advanced Conjugate tricks for higher roots [Optional]
There is a higher degree generalization of the above formula. The generalization is based on the following
two formula:
Formula 3.11.
1. an − bn = (a − b)(an−1 + an−2 b + an−3 b2 + . . . + abn−2 + bn−1 )
2. an + bn = (a + b)(an−1 − an−2 b + an−3 b2 − . . . + abn−2 + bn−1 ) when n is odd.
For example, let us take n = 3. Suppose we see the following problem:
9
Problem 3.12.
lim √
3
x→12
x − 12
x + 15 − 3
Recall that we have (a − b)(a2 + ab + b2 ) = a3 − b3 . So we multiply on both numerator and denominator
√
√
2
by 3 x + 15 + 3 3 x + 15 + 9. We would have:
x − 12
x + 15 − 3
√
√
2
(x − 12)( 3 x + 15 + 3 3 x + 15 + 9)
= lim √
√
√
x→12 ( 3 x + 15 − 3)( 3 x + 152 + 3 3 x + 15 + 9)
√
√
2
(x − 12)( 3 x + 15 + 3 3 x + 15 + 9)
= lim
x→12
(x + 15) − 27
√
√
2
3
(x − 12)( x + 15 + 3 3 x + 15 + 9)
= lim
x→12
x − 12
√
√
2
3
x + 15 + 3 3 x + 15 + 9
= lim
lim √
3
x→12
x→12
=21
Here we can plug in the value x = 12 because the last expression is continuous at x = 0.
3.6
Exercises
Problem 3.13.
3x4 − 8x3 + 2x + 1
x→∞
x5 + x4 + 8
lim
Problem 3.14.
x6 − 7
x→∞ 7x6 + 5x4 + 3x2 + 1
lim
Problem 3.15.
5ex + 8x2 − 2 sin x
x→∞
4ex − 3x + 4
lim
Problem 3.16 (HARD).
4 tan x +
limπ
x→ 2
Problem 3.17.
9 tan x +
3
x− π
2
8
π
x− 2
sin(3x) sin2 (5x)
x→0
sin(4x)x2
lim
Problem 3.18.
sin(e2x ) sin2 (x)
x→−∞
sin(ex )ex
lim
Problem 3.19.
cot2 (2x) + 1
x→0 sin−2 (3x)
lim
Problem 3.20.
lim √
x→1
x2 − 1
x+8−3
10
Problem 3.21 (HARD).
√
lim
x→0
4
cos x + 3 − 2
x2
Substitution [Optional]
4.1
Linear subs [Optional]
The most common type of substitution is linear substitution. Namely, when you see a limit in terms of a
variable x, you can let t = ax + b for some constant a, b, and then transform everything in terms of b. In
particular, we have the following fact:
Fact 4.1. Let t = ax + b. Then we have the following:
lim f (x) =
x→p
1
lim f ( (t − b))
t→ap+b
a
In general, suppose we let t = g(x) for some invertible function g, then we should have the following fact:
Fact 4.2. Let t = g(x) for some invertible function g, and let g −1 denote its inverse. Then we have the
following:
lim f (x) = lim f (g −1 (t))
x→p
4.2
t→g(p)
Sub to change the limit [Optional]
Sometimes, it might happen that you see a limit that you don’t like. For example, when we see lots of
trig functions, we want to change all trig functions into sin x and try to use the fact that limx→0 sinx x = 1.
However, this might be awkward if we don’t have x → 0. For example, think about the following problem:
Problem 4.3.
lim
x→ π
2
cos x
x − π2
When we see a limit like this, we would like to change all cos x into sin x. But then we would like to have
x → 0 so that we can use the formula limx→0 sinx x = 1.
So the key here is to find “something” other than x that goes to 0, and we substitute all x by this
“something” instead. What would go to 0? Well, if x goes to a value a, then x − a will always go to 0. So
we do the substitution by letting t = x − π2 in our case. Then we would have t = 0, and we would substitute
x by t + π2 everywhere. So we have:
cos x
x − π2
cos(t + π2 )
= lim
t→0 (t + π ) − π
2
2
cos(t + π2 )
= lim
t→0
t
− sin(t)
= lim
t→0
t
=−1
lim
x→ π
2
Here are some common way of substitutions and the corresponding change in limit.
Fact 4.4.
1. If x → p and t = x − p, then t → 0.
11
2. If x → ∞ and t = x − a, then t → ∞.
3. If x → ∞ and t = kx for some k > 0, then t → ∞.
4. If x → ∞ and t = kx for some k < 0, then t → −∞.
5. If x → ∞ and t =
x → ∞)
1
x,
then t → 0+ . (The one sided limit as t → 0+ will be the same as the limit
6. If x → ∞ and t = ex , then t → ∞.
7. If x → ∞ and t = ln x, then t → ∞.
8. If x → −∞ and t = ex , then t → 0+ .
9. If x → 0 and t = ln x, then t → −∞.
10. If x → 0+ and t = x1 , then t → ∞.
11. If x → 0− and t = x1 , then t → −∞.
4.3
Sub to change ugly expression [Optional]
Now another use of substitution is to get rid of ugly expression you hate. As a simple example, let us look
at this problem.
Problem 4.5.
sin(ln x)
ln x
Now, we would really like to see sin x instead of sin(ln x). So as a result, we can let t = ln x and do the
substitution. Then we would have:
lim
x→1
sin(ln x)
ln x
sin t
= lim
t→0 t
=1
lim
x→1
Another example is to get rid of square roots. For example, let us look at problem 3.10 again.
Problem 4.6.
√
x+9−3
2x
√
√
The expression we really hate is x + 9. So we let t = x + 9, (and then we would have x = t2 − 9).
Then we would have the following.
lim
x→0
√
x+9−3
2x
t−3
= lim
t→3 2(t2 − 9)
t−3
= lim
t→3 2(t + 3)(t − 3)
1
= lim
t→3 2(t + 3)
1
=
12
lim
x→0
12
The moral of the story is this: whenever you see something you hate, simply substitute it away.
4.4
Universal trig formula [Optional]
The universal trig formula refers to the formula 3.9. Namely, if we let t = tan( x2 ), then all trig functions in
x can be expressed as a “polynomial over polynomial” type thing. And this is really handy, since we can
then use the division trick as introduced in section 3.1.
For example, you may see the following problem.
Problem 4.7.
lim
x→ π
2
cos x
1 − tan x2
Then we can do this by setting t = tan x2. With this substitution, we shall have:
lim
x→ π
2
cos x
1 − tan( x2 )
(t2 − 1)/(t2 + 1)
t→1
1−t
t+1
= lim 2
t→1 t + 1
=1
= lim
4.5
Exercises
Try to solve the following problem using substitution.
Problem 4.8.
limπ
sin x − 1
x − π2
lim3π
sin x + 1
2
(x − 3π
2 )
x→ 2
Problem 4.9.
x→
Problem 4.10.
2
sin(ex ) + 1
3π 2
x→(ln(3π)−ln 2) (ex −
2 )
lim
Problem 4.11.
1
lim (x sin( ))
x→∞
x
Problem 4.12.
lim q
x→∞
5
5.1
1
x
9+
1
x
−3
Sandwich Theorem
How to find a good sandwich
Let us take a good look at an example first.
Problem 5.1.
1
lim x3 sin( )
x
x→0
13
Now, the correct sandwich for this problem is the following inequality.
1
−|x3 | ≤ x3 sin( ) ≤ |x3 |
x
By taking limit of x → 0, we see that both sides of the inequality would go to 0. So we see that
limx→0 x3 sin( x1 ) = 0.
The reason why this sandwich works is that sin(blah) always takes value between −1 and 1. A bounded
function is like a tamed little sheep, and they always get pushed around by other functions.
In particular, we have the following fact:
Fact 5.2. For limit of the following two forms:
lim
blah · sin(blee)
lim
blah · cos(blee)
x→somewhere
x→somewhere
We should always use the following sandwiches:
−|blah| ≤ blah · sin(blee) ≤ |blah|
−|blah| ≤ blah · cos(blee) ≤ |blah|
To take things furthrer, we see that we don’t really need to have sin or cos. All we need is a function
bounded between −1 and 1.
Fact 5.3. Let f (x) be any function, and suppose g(x) is a function always bounded between −1 and 1. For
limit of the following form:
lim
f (x) · g(x)
x→somewhere
We should always use the following sandwich:
−|f (x)| ≤ f (x) · g(x) ≤ |f (x)|
As an example, let us consider the following problem:
Problem 5.4.
lim (x − 3)
x→3
ex − e3
|ex − e3 |
To solve this, we first observe that the function g(x) =
we let f (x) = x − 3, and do the following sandwich:
−|x − 3| ≤ (x − 3)
ex −e3
|ex −e3 |
is always bounded between −1 and 1. So
ex − e3
≤ |x − 3|
|ex − e3 |
And by sandwich theorem we can see that the limit we want to solve is 0.
14
5.2
Advanced sandwich [Optional]
Sometimes it might happen that although we have the form f (x) · g(x), and although g(x) is bounded, we
have g(x) bounded between two number a, b instead of −1, 1. What should we do in this case? Let us first
look at the following problem.
Problem 5.5.
1
lim (x − 4)2cos( x−4 )
x→4
Now, we observe that since −1 ≤ cos(blah) ≤ 1, we must have 2−1 ≤ 2cos(blah) ≤ 21 . So we let
1
g(x) = 2cos( x−4 ) , a bounded function, and let f (x) = x − 4, the function that we shall put into absolute
value and used as the sandwich. But now we in fact have two situations.
1
Suppose x − 4 ≥ 0. Then we shall have 21 |x − 4| ≤ (x − 4)2cos( x−4 ) ≤ 2|x − 4|. On the other hand, if we
1
have x − 4 ≤ 0, we would have −2|x − 4| ≤ (x − 4)2cos( x−4 ) ≤ − 12 |x − 4|. To accomodate both situations at
once, we should eventually choose to use the following sandwich:
1
−2|x − 4| ≤ (x − 4)2cos( x−4 ) ≤ 2|x − 4|
And with this sandwich we can see that the limit we want to compute is 0.
To sum up, we have the following strategy:
Fact 5.6. Let g(x) be a bounded function such that a ≤ g(x) ≤ b, and let f (x) be any function. Suppose we
want to compute the following limit:
lim
f (x) · g(x)
x→something
Then if |a| ≥ |b|, we shall use the following sandwich:
−|a| · |f (x)| ≤ f (x) · g(x) ≤ |a| · |f (x)|
If |a| ≤ |b|, we shall use the following sandwich:
−|b| · |f (x)| ≤ f (x) · g(x) ≤ |b| · |f (x)|
5.3
Sandwich to infinity [Optional]
By sandwich to infinity, we mean to compare the growth rate of different functions. This arises in many
different context, for example, when you try to do the advanced division trick as in section 3.2.
All these advanced sandwiches will be determined by inequalities between functions. And the best way
to “measure” the growth rate of a function is to compare it with the power functions. Let us start with a
definition first:
(x)
. If L = ∞, then we
Definition 5.7. For two functions f (x), g(x), we consider the limit L = limx→a ] fg(x)
say f (x) >> g(x) at x = a. If L = 0, we say f (x) << g(x) at x = a. If 0 < L < ∞, then we say f ∼ g at
x = a.
Let us first talk about growth rate when x → ∞.
Fact 5.8.
1. xa << xb when a < b.
2. ex >> xa for all a.
3. x0 << ln x << xa for all a > 0.
15
4. sin x << xa for all a > 0.
5. cos x << xa for all a > 0.
6. ax >> bx for a > b.
Here is a list of growth rate at x = 0.
Fact 5.9.
1. xa << xb when a > b.
2. x0 << ln x << x−a for all a > 0.
3. sin x ∼ x.
4. cos x ∼ 1.
Remember these growth rates will help you tremendously in finding good sandwiches.
5.4
Exercises
Problem 5.10.
p
lim ( x2 − 9 − 4)ecos(ln x)
x→5
Problem 5.11.
lim (ex − x)
x→∞
Problem 5.12.
sin x
2
lim ln( x)
π | sin x|
x→ π
2
Problem 5.13 (HARD).
lim x ln x
x→0
Problem 5.14 (HARD).
lim xx
x→0+
Problem 5.15 (HARD).
1
lim x x
x→0+
Problem 5.16 (HARD).
1
lim x x
x→∞
16