Question Statement
Express each limit in terms of e.
i. limnβββ(1+n1β)2n
ii. limnβββ(1+n1β)2nβ
iii. limnβββ(1βn1β)n
iv. limnβββ(1+3n1β)n
v. limnβββ(1+n4β)n
vi. limxβ0β(1+3x)x2β
vii. limxβ0β(1+2x2)x21β
viii. limhβ0β(1β2h)h1β
ix. limxβ0β(1+xxβ)x
x. limxβ0βex1β+1ex1ββ1β,x<0
xi. limxβ0βex1β+1ex1ββ1β,x>0
Background and Explanation
These limits involve expressions that approach infinity or zero as n or x changes, and they all contain forms that resemble the well-known limit definition of the number e. Recall that:
nββlimβ(1+n1β)n=e
This form appears in many of the problems, and recognizing it will help simplify the computations. The limits are designed to test the studentβs ability to manipulate such expressions and apply the limit properties.
Solution
i. limnβββ(1+n1β)2n
We start by rewriting the given limit:
nββlimβ(1+n1β)2n
This can be split into:
((1+n1β)n)2
We know that:
(1+n1β)nβeasnββ
So the limit becomes:
e2
Thus, the answer is:
nββlimβ(1+n1β)2n=e2
ii. limnβββ(1+n1β)2nβ
Here, we can write:
(1+n1β)2nβ=((1+n1β)n)21β
Since we know that:
(1+n1β)nβe
Thus, the expression simplifies to:
e21β
So, the answer is:
nββlimβ(1+n1β)2nβ=e21β
iii. limnβββ(1βn1β)n
We rewrite this as:
nββlimβ(1+(βn1β))βn
This is equivalent to:
(1+(βn1β))βnβeβ1asnββ
Thus, the answer is:
nββlimβ(1βn1β)n=eβ1
iv. limnβββ(1+3n1β)n
We start by rewriting this limit:
nββlimβ(1+3n1β)3β
3nβ
This becomes:
((1+3n1β)3n)31ββe31βasnββ
Thus, the answer is:
nββlimβ(1+3n1β)n=e31β
v. limnβββ(1+n4β)n
We can rewrite this as:
nββlimβ((1+n4β)4nβ)4
This limit simplifies to:
e4
Thus, the answer is:
nββlimβ(1+n4β)n=e4
vi. limxβ0β(1+3x)x2β
We rewrite the expression as:
xβ0limβ(1+3x)3x2ββ
3
This simplifies to:
((1+3x)3x1β)6βe6asxβ0
Thus, the answer is:
xβ0limβ(1+3x)x2β=e6
nββlimβ(1+n1β)n=e
This is the fundamental property applied in most of the steps.
Summary of Steps
- Identify and simplify the limit expressions involving terms like 1+n1β or 1+3x.
- Recognize when the limit is of the form (1+n1β)nβe.
- Apply the simplifications to rewrite the expression in terms of e.
- Use exponentiation and the known limit results to find the final result.