Simple Limit Problem...Please Help!?

Question

1. lim(1+1/n)^n
n--> infinity

2. lim(sinx)/x
x--> 0

I need explinations and steps! this would help alot!! Thank You!!

Answer 1

1. If you expand (1+1/n)^n and take the limit as n approaches infinity, you get the Taylor series for e^x when x=1.

2. Again, use Taylor series. The Taylor series of sin(x) is given by x - x^3/3! + x^5/5! - ... Dividing by x and subsequently evaluating as x goes to 0, it is clear that sin(0)/0 approaches 1.

I'm not sure if you have encountered Taylor series yet, but this is the simplest way to approach this problem. A different method for solving part 2 would be to use a geometric argument, which is hard for me to explain since I can't draw anything. While there are other ways to prove 1, I do not know of any simpler way than the solution involving the Taylor series for e^x.

Answer 2

1.
Expanding (1 + 1/n)^n by the binomial theorem:
(1 + 1/n)^n
= (1 + n(1/n) + ( n(n - 1)/2! )(1/n)^2
+ ( n(n - 1)(n - 2)/3! )(1/n)^3 + ... )
= 1 + 1 + 1(1 - 1/n)2! + 1(1 - 1/n)(1 - 2/n)3! + ...
As n -> infty, this becomes:
1 + 1 + 1/2! + 1/3! + ...
= e.

2.
lim(x -> 0) sin(x) / x
Expand sin(x) as a power series:
lim(x -> 0) (x - x^3 / 3! + x^5 / 5! - ...) / x
= lim(x -> 0) (1 - x^2 / 3! + x^4 / 5! - ...)
= 1.

Answer 3

1. By definition it is equal to e (2.71828...)

2. Derivative of sin x / derivative of x

(Cos 0) /1 = 1
.

Answer 4

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